Newest questions tagged precipitation - Chemistry Stack Exchange most recent 30 from chemistry.stackexchange.com 2019-08-20T07:15:48Z https://chemistry.stackexchange.com/feeds/tag?tagnames=precipitation&sort=newest http://www.creativecommons.org/licenses/by-sa/3.0/rdf https://chemistry.stackexchange.com/q/119311 -1 pH and precipitation of cations with H2S solution [on hold] Abdullah Saad https://chemistry.stackexchange.com/users/71263 2019-08-17T17:26:47Z 2019-08-17T17:54:31Z <p>In a solution, both <span class="math-container">$\ce{Pb^2+}$</span> and <span class="math-container">$\ce{Zn^2+}$</span> ion has a molarity of .01 M . In which pH only the <span class="math-container">$\ce{Pb^2+}$</span> will precipitate but <span class="math-container">$\ce{Zn^2+}$</span> will not do so? The solubility for <span class="math-container">$\ce{H2S}$</span> and <span class="math-container">$\ce{ZnS}$</span> is given <span class="math-container">$\mathrm{1.1×10^{-22}}$</span> and <span class="math-container">$\mathrm{1×10^{-21}}$</span> respectively.</p> <p>This is the problem. But I don't understand how to relate those things with each other. Can anybody explain?</p> https://chemistry.stackexchange.com/q/117898 0 Which metal ions precipitate in NH4OH? [closed] Veritas https://chemistry.stackexchange.com/users/79984 2019-07-11T00:01:45Z 2019-07-11T21:32:42Z <p>From <a href="https://en.wikipedia.org/wiki/Metal_ammine_complex" rel="nofollow noreferrer">Wikipedia-1</a>, and from an answer here on <a href="https://chemistry.stackexchange.com/questions/68139/which-metals-hydroxides-dissolve-in-excess-nh4oh">Stack-Exchange</a>, I can list at least the following elements which will dissolve in aqueous <span class="math-container">$\ce{NH4OH}$</span>:</p> <ul> <li><span class="math-container">$\ce{Cr(III), Co(III), Ni(II), Cu(I, II), Zn(II), Ag(I),}$</span> along with "several platinum group metals"</li> </ul> <p>From reading <a href="https://en.wikipedia.org/wiki/Ammonia#Liquid_ammonia_as_a_solvent" rel="nofollow noreferrer">Wikipedia-2</a>, I also suspect the following elements to dissolve in liquid ammonia:</p> <ul> <li><span class="math-container">$\ce{Mg, Ca, Sr, Ba, Eu, Yb, Li, Na, K}$</span> and <span class="math-container">$\ce{Rb}$</span></li> </ul> <p>As a matter of fact, according to Wikipedia "almost all metal ions bind ammonia as a ligand." I thought it would be interesting, and potentially very useful, to find out which metal ions are NOT soluble in <span class="math-container">$\ce{NH4OH}$</span>.</p> <p>I can list two for now: </p> <ul> <li><span class="math-container">$\ce{Fe(OH)3}$</span> </li> <li><span class="math-container">$\ce{Mg(OH)2}$</span> with a solubility product of only <span class="math-container">$5.61×10^{−12}$</span> (credit to Oscar Lanzi)</li> </ul> https://chemistry.stackexchange.com/q/116988 1 Which salts can be used to precipitate chloride ions? [closed] user65414 https://chemistry.stackexchange.com/users/65414 2019-06-17T18:42:45Z 2019-06-19T08:09:43Z <p>We don’t have AgNO3 in the lab. I have no idea what else can be useful. I am washing a product and I want to make sure the chlorides are washed away.</p> https://chemistry.stackexchange.com/q/116724 1 Can I safely use a separatory funnel to extract the precipitate from a solution? Veritas https://chemistry.stackexchange.com/users/79984 2019-06-12T08:12:56Z 2019-06-12T14:48:45Z <p>Can a separatory funnel be used to collect a precipitate, or would that risk clogging the stopcock on it?</p> <p>I'd like to recoup the precipitate for filtering without passing the entire solution through a filter (to avoid contaminating it, and also for convenience / saving time).</p> https://chemistry.stackexchange.com/q/115845 0 Why does Mg(HCO₃)₂ on heating gives Mg(OH)₂? [duplicate] user226375 https://chemistry.stackexchange.com/users/76375 2019-05-24T04:01:39Z 2019-05-24T17:05:58Z <div class="question-status question-originals-of-duplicate"> <p>This question already has an answer here:</p> <ul> <li> <a href="/questions/65901/thermal-decomposition-of-magnesium-bicarbonate" dir="ltr">Thermal decomposition of magnesium bicarbonate</a> <span class="question-originals-answer-count"> 2 answers </span> </li> </ul> </div> <p>Normally, bicarbonates on heating forms carbonates. Why does Mg not give this too?</p> <p>In a reference book I am referring to, it is given that it is due to high solubility product of Mg(OH)₂ compared to Mg(CO₃) .</p> <p>Then shouldn't the latter form? As even small amounts of it will lead to precipitation, driving the equilibrium forward?</p> <p>Edit: This is related to removing temporary hardness of water.</p> https://chemistry.stackexchange.com/q/115745 2 How to find the concentration of barium ions in a mixture of barium sulfate and barium thiosulfate? valkedin https://chemistry.stackexchange.com/users/79443 2019-05-22T11:17:11Z 2019-05-22T15:12:45Z <p>I have a hard time with selective precipitation problem.</p> <blockquote> <p>A mixture of <span class="math-container">$\ce{BaSO4}$</span> and <span class="math-container">$\ce{BaS2O3}$</span> is shaken with pure water until a saturated solution is formed. Both solids remain in excess. What is <span class="math-container">$[\ce{Ba^{2+}}]$</span> in the saturated solution? <span class="math-container">$K_\mathrm{sp}(\ce{BaSO4}) = 9 \times 10^{-11}$</span> and <span class="math-container">$K_\mathrm{sp}(\ce{BaS2O3}) = 4 \times 10^{-10}$</span>.</p> </blockquote> <p>My solution: </p> <p>Because <span class="math-container">$K_\mathrm{sp}$</span> of <span class="math-container">$\ce{BaSO4} &lt; \ce{BaS2O3}$</span>, <span class="math-container">$\ce{BaS2O3}$</span> will dissolved first. When <span class="math-container">$\ce{BaS_2O_3}$</span> dissolved, the concentration of <span class="math-container">$[\ce{Ba^{2+}}] = [\ce{S2O3^2-}] = \sqrt{K_\mathrm{sp}(\ce{BaS2O3})} = 2 \times 10^{-5}$</span>.</p> <p>That means, in the solution, there are:</p> <p><span class="math-container">\begin{align} [\ce{Ba^2+}] &amp;= 2 \times 10^{-5}\\ [\ce{S2O3^2-}] &amp;= 2 \times 10^{-5}\\ [\ce{SO4^2-}] &amp;= \frac{K_\mathrm{sp}(\ce{BaSO4})}{[\ce{Ba^2+}]} = \frac{9 \times 10^{-11}}{2 \times 10^{-5}} = 4.5 \times 10^{-6} \end{align}</span></p> <p>But the solution is <span class="math-container">$\ce{[Ba^2+]} = \sqrt{4.9 \times 10^{-10}}$</span>.</p> <p>What have I missed?</p> <p>EDIT 1 : <span class="math-container">\begin{align} \ce{BaSO4 &amp;&lt;=&gt; Ba^2+ + SO4^2- }\\ \ce{BaS2O3 &amp;&lt;=&gt; Ba^2+ + S2O3^2- } \end{align}</span></p> <p>The Solubility Product Equation : </p> <p><span class="math-container">\begin{align} K_\mathrm{sp}(\ce{BaSO4}) &amp;= [\ce{Ba^2+}] \times [\ce{SO4^2-}] \\ K_\mathrm{sp}(\ce{BaS2O3}) &amp;= [\ce{Ba^2+}] \times [\ce{S2O3^2-}] \end{align}</span></p> <p>When the solids dissolves, it will produce <span class="math-container">$\ce{Ba^2+}$</span> ion, <span class="math-container">$\ce{SO4^2-}$</span> ion and <span class="math-container">$\ce{S2O3^2-}$</span> ion.</p> <p><span class="math-container">$\ce{BaSO4}$</span> will stop dissolving when <span class="math-container">$\ce{Ba^2+}$</span> ion concentration in solution when multiply with <span class="math-container">$\ce{SO4^2-}$</span> ion concentration is equal to solubility product of <span class="math-container">$\ce{BaSO4}$</span>, which is <span class="math-container">$K_\mathrm{sp}(\ce{BaSO4}) = 9 \times 10^{-11}$</span></p> <p>And at the same time, </p> <p><span class="math-container">$\ce{BaS2O3}$</span> will stop dissolving when <span class="math-container">$\ce{Ba^2+}$</span> ion concentration in solution when multiply with <span class="math-container">$\ce{S2O3^2-}$</span> ion concentration is equal to solubility product of <span class="math-container">$\ce{BaS2O3}$</span>, which is <span class="math-container">$K_\mathrm{sp}(\ce{BaS2O3}) = 4 \times 10^{-10}$</span></p> <p>Solubility of <span class="math-container">$\ce{BaSO4}$</span> in water is <span class="math-container">\begin{align} s &amp;= \sqrt{K_\mathrm{sp}(\ce{BaSO4})}\\ s &amp;= \sqrt{9 \times 10^{-11}} \\ s &amp;= 3 \times 10^{-5.5} \end{align}</span></p> <p>Solubility of <span class="math-container">$\ce{BaS2O3}$</span> in water is <span class="math-container">\begin{align} s &amp;= \sqrt{K_\mathrm{sp}(\ce{BaS2O3})}\\ s &amp;= \sqrt{4 \times 10^{-10}}\\ s &amp;= 2 \times 10^{-5} \end{align}</span></p> <p>Now, when the solids is mixed in water, There should be <span class="math-container">$[\ce{SO4^2-}] =3 \times 10^{-5.5}$</span> and <span class="math-container">$[\ce{S2O3^2-}] =2 \times 10^{-5}$</span>.</p> <p>Now I confuse what ion concentration should I use to determine <span class="math-container">$[\ce{Ba^2+}]$</span></p> <p>EDIT 2:</p> <p><span class="math-container">\begin{align} K_\mathrm{sp}(\ce{BaSO4}) &amp;= [\ce{Ba^2+}] \times [\ce{SO4^2-}] \\ \frac{K_\mathrm{sp}(\ce{BaSO4})}{[\ce{Ba^2+}]} &amp;= [\ce{SO4^2-}] \\ \end{align}</span> and <span class="math-container">\begin{align} K_\mathrm{sp}(\ce{BaS2O3}) &amp;= [\ce{Ba^2+}] \times [\ce{S2O3^2-}] \\ \frac{K_\mathrm{sp}(\ce{BaS2O3})}{[\ce{Ba^2+}]} &amp;= [\ce{S2O3^2-}] \\ \end{align}</span></p> <p>From <a href="https://chemistry.stackexchange.com/a/115750/79443">@santimirandarp answer</a>, I got that : </p> <p><span class="math-container">\begin{align} [\ce{Ba^2+}] &amp;= [\ce{S2O3^2-}] + [\ce{SO4^2-}]\\ [\ce{Ba^2+}] &amp;= \frac{K_\mathrm{sp}(\ce{BaS2O3})}{[\ce{Ba^2+}]} + \frac{K_\mathrm{sp}(\ce{BaSO4})}{[\ce{Ba^2+}]}\\ [\ce{Ba^2+}] &amp;= \frac{4 \times 10^{-10}}{[\ce{Ba^2+}]} + \frac{9 \times 10^{-11}}{[\ce{Ba^2+}]}\\ [\ce{Ba^2+}] &amp;= \frac{4.9 \times 10^{-10}}{[\ce{Ba^2+}]} \\ [\ce{Ba^2+}]^2 &amp;= 4.9 \times 10^{-10} \\ [\ce{Ba^2+}] &amp;= \sqrt{4.9 \times 10^{-10}} \\ \end{align}</span></p> <p>My problem solved. Thank You </p> https://chemistry.stackexchange.com/q/115225 4 CaSO4 precipitate in mixture of Na2SO4 and CaCl2 solutions Dumbstudent https://chemistry.stackexchange.com/users/79083 2019-05-13T06:19:43Z 2019-05-13T18:18:44Z <p>I have been stuck on a chemistry problem for a long time now and if anyone here can help me I would be eternally grateful.</p> <blockquote> <p><strong>40.</strong> Du tillsätter <span class="math-container">$\pu{100 ml}$</span> <span class="math-container">$\pu{0,01 M}$</span> <span class="math-container">$\ce{Na2SO4}$</span>-lösning i <span class="math-container">$\pu{100 ml}$</span> <span class="math-container">$\pu{0,02 M}$</span> <span class="math-container">$\ce{CaCl2}$</span>-lösning. Det bildas en <span class="math-container">$\ce{CaSO4}$</span>-fällning. Hur mycket av <span class="math-container">$\pu{0,01 M}$</span> <span class="math-container">$\ce{Na2SO4}$</span>-lösning måste ännu tillsättas för att den bildade <span class="math-container">$\ce{CaSO4}$</span>-fällningen just och just ska upplösas helt? <span class="math-container">$K_\mathrm{s}(\ce{CaSO4}) = \pu{2,5e-5 M}$</span>. <span class="math-container">$(t = \pu{25 °C})$</span> (2p)</p> <p>a. <span class="math-container">$\pu{0,43 l}$</span><br> b. <span class="math-container">$\pu{0,48 l}$</span><br> c. <span class="math-container">$\pu{0,53 l}$</span><br> d. <span class="math-container">$\pu{0,58 l}$</span> </p> </blockquote> <p>Translation: We add 100 ml of 0.01 M <span class="math-container">$\ce{Na2SO4}$</span>-solution to 100 ml of 0.02 M <span class="math-container">$\ce{CaCl2}$</span>-solution. A precipitate of <span class="math-container">$\ce{CaSO4}$</span> is formed. What is the (minimal) volume of 0.01 M <span class="math-container">$\ce{Na2SO4}$</span>-solution that needs to be added to the mix for the <span class="math-container">$\ce{CaSO4}$</span> precipitate to be just dissolved completely?</p> <p>The problem then lists the solubility constant, <span class="math-container">$K_\mathrm{s}$</span>, for <span class="math-container">$\ce{CaSO4}$</span> as <span class="math-container">$2.5\cdot 10^{-5}$</span>.</p> <p>The only way I can think of solving this myself is by looking at the added <span class="math-container">$\ce{Na2SO4}$</span>-solution as just water and using the <span class="math-container">$K_\mathrm{s}$</span> value for <span class="math-container">$\ce{CaSO4}$</span> to see what concentration of <span class="math-container">$\ce{CaSO4}$</span> is possible in just water. Then calculate the amount of moles of precipitate that was actually formed to see how much water I would need to add to the mixture:</p> <p><span class="math-container">$$K_\mathrm{s}(\ce{CaSO4}) = \pu{2.5e-5 M^2} = [\ce{Ca^2+}][\ce{SO4^2-}]$$</span></p> <p><span class="math-container">$$\therefore [\ce{Ca^2+}] = [\ce{SO4^2-}] = \sqrt{\pu{2.5e-5 M^2}} = \pu{5e-3 M}$$</span></p> <p><span class="math-container">$$C = \frac{n}{V} \to n = CV$$</span></p> <p>Amount of precipitate formed:</p> <p><span class="math-container">$$n(\ce{CaSO4}) = \pu{0.02 M} \cdot \pu{0.1 dm^3} = \pu{0.002 mol}$$</span></p> <p>Required total volume of water to dissolve <span class="math-container">$\pu{0.002 mol}$</span> of <span class="math-container">$\ce{CaSO4}$</span>: </p> <p><span class="math-container">$$V = \frac{n}{C} = \frac{\pu{0.002 mol}}{\pu{5e-3 M}} = \pu{0.4 dm^3}$$</span></p> <p>This is wrong, but I can't figure out the right way to go about this. The answer surely has to do with how well <span class="math-container">$\ce{CaSO4}$</span> is dissolved in a <span class="math-container">$\ce{Na2SO4}$</span>-solution as opposed to just water.</p> https://chemistry.stackexchange.com/q/114331 2 Cleaning a stubborn Electrospray residue AnonymousGraduateStudent https://chemistry.stackexchange.com/users/77109 2019-04-25T17:51:15Z 2019-04-25T17:51:15Z <p>I have an electrospray ionization capillary tip that sprays into a chamber that contains some circuit boards. An aqueous CsI sample is apparently causing a white residue to build up on these boards after some time. I thought the residue was the salt itself, but I can't dissolve it with water. I tried methanol and that didn't work either. Does anyone know what this could be and how to clean it? I'm worried about excess build up over time.</p> <p>I considered that it might be CsOH, but that should also rinse away with water because its so soluble.</p> https://chemistry.stackexchange.com/q/112818 0 Precipitation of Cu2+ and Pb2+ ions Trey Anupong https://chemistry.stackexchange.com/users/76916 2019-04-15T19:20:02Z 2019-04-15T19:30:59Z <p>Hi chemistry enthusiasts! I am working on my scholarship exam practice. I believe this exam assumes high school + first year university knowledge although the standard may be varied across the world. And I'm not quite sure what I did wrong. Could you please have a look?</p> <blockquote> <p>There is an aqueous solution containing Cu2+ and Pb2+ ions. The most suitable reagent to precipitate one of the two ions from the solution is (1) nitric acid (2) sodium carbonate (3) sulfuric acid (4) hydrogen sulfide</p> </blockquote> <p>I picked answer (4) hydrogen sulfide because I think that it will react with copper (II) ions and will precipitate out. However, the correct answer is (3) sulfuric acid. How is this answer correct and what did I do wrong for hydrogen sulfide. Your advice will be much appreciated!</p> https://chemistry.stackexchange.com/q/112772 1 Are precipitations exothermic and/or endothermic? chompion https://chemistry.stackexchange.com/users/75961 2019-04-15T04:54:13Z 2019-05-17T05:19:47Z <p>Should be an easy one. I'm fumbling a concept. I've read precipitations are exothermic. Is this accurate? Why would there be no endothermic precipitation reactions?</p> https://chemistry.stackexchange.com/q/112762 1 Making hard water Jason Rusk https://chemistry.stackexchange.com/users/76994 2019-04-15T01:38:47Z 2019-05-15T17:02:20Z <p>Does anyone know how to make hard water with a hardness of 200+ ppm in terms of <span class="math-container">$\ce{CaCO3}$</span>?</p> <p>I live in Seattle and we have very soft water and we need hard water for testing purposes. Obviously you cannot dissolve <span class="math-container">$\ce{CaCO3}$</span> in water. I've read many "recipes" even ones that claimed you can dissolve <span class="math-container">$\ce{CaCO3}$</span> with a soda stream pop maker. Best I could do was a 60 ppm.</p> <p>I've tried baking soda and Epsom salt and was able to make 130 ppm.</p> <p>I've tried <span class="math-container">$\ce{HCl}$</span> to dissolve it in water, but the resulting pH was not suitable and if I tried to raise the pH, the <span class="math-container">$\ce{CaCO3}$</span> precipitated back out of the water.</p> <p>I'm at a loss on a "recipe" that actually works. Am I trying to replicate the possible? Any ideas?</p> https://chemistry.stackexchange.com/q/112647 0 Influence of pH on precipitate crystallite sizes? Roam https://chemistry.stackexchange.com/users/76932 2019-04-13T07:17:58Z 2019-04-13T12:45:56Z <p>It's normally assumed that a higher pH of solution would slow nucleation rates and hence lead to larger crystal formation. I've found literature that it is due to solid-liquid inter-facial tension. Which is a concept that I do not fully understand.</p> <p>In my experiments, higher pH solutions led to the formation of smaller crystallites. I can't make sense of why its contradicting the literature. Here are the images from my experiments:</p> <p><a href="https://i.stack.imgur.com/TBPmK.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/TBPmK.jpg" alt="Images of crystals at different pH"></a></p> <p>My samples are precipitates produced from mixing aqueous <span class="math-container">$\ce{CaCl2}$</span> +<span class="math-container">$\ce{MgCl2}$</span> and <span class="math-container">$\ce{Na2CO3}$</span> solutions. The elevated pH images were produced through the addition <span class="math-container">$\ce{NaOH}$</span> to the previously mentioned mixture.</p> <p>If you have any comments or ideas, it would be greatly appreciated. I am a engineering student who is trying his hand at inorganic chemistry. </p> <p>Thanks in advance.</p> https://chemistry.stackexchange.com/q/111790 0 Eutectic Phase Diagram of Al-Si Analysis Noam Azulay https://chemistry.stackexchange.com/users/76364 2019-03-30T10:53:32Z 2019-03-30T16:04:50Z <p><a href="https://i.stack.imgur.com/xLBiu.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/xLBiu.png" alt="enter image description here"></a></p> <p>Hello, How can I know from this phase diagram: 1. The composition of the alloy Al-Si that allows precipitation to occure? 2. The range of temperatures where the Al-Si is soluble? 3. The range of temperatures in which precipitation can take place?</p> <p>Regarding the 1st and 2nd question- I assume that 1.6%wt Si at 557 degrees Celsius is the max solubility limit of this alloy, and that you can't have precipitation if there are higher percentages of Si in the aluminium. But what about precipitation temperatures? Which conditions allow having precipitation?</p> https://chemistry.stackexchange.com/q/111135 1 How to convert soluble colorless [HgI₄]²⁻ back to insoluble red HgI₂? [closed] HGK https://chemistry.stackexchange.com/users/75858 2019-03-17T18:34:41Z 2019-03-18T05:13:37Z <p>When a solution contains <span class="math-container">$\ce{Hg^2+}$</span> and <span class="math-container">$\ce{I-}$</span>, depend on the concentration ratio of <span class="math-container">$\ce{Hg^2+}$</span> and <span class="math-container">$\ce{I-}$</span>, the following reactions can happen:</p> <p><span class="math-container">$$\ce{Hg^2+ + 2 I- ⟶ \underset{insoluble red color solid}{HgI2}}$$</span></p> <p>If there is excess <span class="math-container">$\ce{I-}$</span>, then this red solid will continue to react with excess <span class="math-container">$\ce{I-}$</span> to form an colorless soluble complex:</p> <p><span class="math-container">$$\ce{HgI2 + I- ⟶ [HgI3]^-}$$</span></p> <p><span class="math-container">$$\ce{[HgI3]^- + I- ⟶ [HgI4]^2-}$$</span></p> <p>Is there a way to revert <span class="math-container">$\ce{[HgI4]^2-}$</span> back to <span class="math-container">$\ce{HgI2}$</span>?</p> https://chemistry.stackexchange.com/q/110542 1 how can I predict the rate of water ions precipitation on a surface? MEng https://chemistry.stackexchange.com/users/75463 2019-03-07T11:59:27Z 2019-03-07T16:48:22Z <p>Consider a titanium pot that contains 1kg water with the following ions:</p> <p><span class="math-container">$[Ca^{2+}]=22000$</span> ppm, <span class="math-container">$[Mg^{2+}]=1500$</span> ppm,<span class="math-container">$[Sr^{2+}]=791$</span> ppm,<span class="math-container">$[Na^{1+}]=48000$</span> ppm,<span class="math-container">$[Cl^{1-}]=120000$</span> ppm,<span class="math-container">$[SO_4^{2-}]=22000$</span> ppm</p> <p>And predict the rate of component precipitation like CaSO4, CaCl2 and ... on the inside wall of the titanium surface at the 20 C, 30C, 50C.</p> <p>Calculate the rate of component precipitation in water not only on the titanium wall.</p> <p>After that, calculate the final steady-state ion contents that exists in the water.</p> https://chemistry.stackexchange.com/q/109695 1 Precipitation of AgCl from the tap water solution of the group 2 chloride Harvey https://chemistry.stackexchange.com/users/74757 2019-02-18T09:41:13Z 2019-02-18T19:03:46Z <p>I had this question about an experiment where the group 2 chloride was dissolved in distilled water and excess <span class="math-container">$\ce{AgNO3}$</span> was added to the solution to form <span class="math-container">$\ce{AgCl}$</span> precipitate. One of the sub-questions read</p> <blockquote> <p>State how the amount of <span class="math-container">$\ce{AgCl}$</span> will change if <strong>tap water</strong> is used to dissolve the group 2 chloride instead of distilled water (will the amount of <span class="math-container">$\ce{AgCl}$</span> formed increase or decrease) and explain why.</p> </blockquote> <p>Can someone please explain how the amount of <span class="math-container">$\ce{AgCl}$</span> formed will change if tap water is used? <em>I assumed that tap water is chlorinated and so the amount of <span class="math-container">$\ce{AgCl}$</span> formed will increase. I don't know if this is right</em>.</p> https://chemistry.stackexchange.com/q/105629 0 Is the water in an aqueous solution undergoing a reaction considered part of the system or surroundings? E. Bauernschmitt https://chemistry.stackexchange.com/users/66527 2018-12-13T23:28:31Z 2018-12-14T11:35:45Z <p><strong>System</strong>: In chemistry, a system is a chemical reaction. A system operates within its surroundings. Energy can move between the two. </p> <p><strong>Surroundings</strong>: The environment around a particular chemical reaction. The chemical reaction is the system.</p> <p>Above are my textbook's definitions for system and surroundings. </p> <p>For example, mixing two aqueous solutions to produce a precipitate. Although the water doesn't undergo any permanent chemical changes, it is still enabling the reaction (much like a catalyst). So is water then considered part of the system?</p> https://chemistry.stackexchange.com/q/103437 1 How to get a precise answer to a textbook precipitation problem? daniel https://chemistry.stackexchange.com/users/5752 2018-10-25T11:53:03Z 2018-10-28T06:04:23Z <p>This question concerns a textbook problem* on precipitation. Briefly, there are <span class="math-container">$2\cdot10^{-4}$</span> of Mn<span class="math-container">$^{2+}$</span> and Cu<span class="math-container">$^{2+}$</span> ions in a liter of solution with <span class="math-container">$0.003$</span> M HClO<span class="math-container">$_4$</span>. This solution is then saturated with H<span class="math-container">$_2$</span>S. The solubility of H<span class="math-container">$_2$</span>S is given as 0.1 mol/L and assumed independent of the presence of other substances in solution. K<span class="math-container">$_{sp}$</span> for MnS is <span class="math-container">$3\cdot 10^{-14}$</span>, K<span class="math-container">$_{sp}$</span> for CuS is <span class="math-container">$8\cdot 10^{-37}.$</span> K<span class="math-container">$_1$</span> and K<span class="math-container">$_2$</span> for H<span class="math-container">$_2$</span>S are given as <span class="math-container">$1.0\cdot 10^{-7}$</span> and <span class="math-container">$1.2\cdot 10^{-13}$</span> respectively. </p> <p>In part one of the question it is found that Mn remains in solution and Cu<span class="math-container">$^{2+}$</span> will precipitate because the product [Cu][S] = <span class="math-container">$2.6\cdot 10^{-20}$</span> far exceeds K<span class="math-container">$_{sp}$</span> of CuS. In part two the problem is to determine how much Cu remains in solution. </p> <p>The solution begins by stating that "most of the Cu<span class="math-container">$^{2+}$</span> will precipitate" and explains that a corresponding increment of [H<span class="math-container">$^+$</span>] will be added to the solution from the H<span class="math-container">$_2$</span>S which has lost S to precipitate, requiring a correction which leads to <span class="math-container">$8\cdot 10^{-21}$</span> moles Cu remaining in solution vs. <span class="math-container">$6\cdot 10^{-21}$</span> moles without the correction.</p> <p>To make the correction the text assumes <em>all</em> the Cu precipitates, so here is my question. </p> <p>The question being asked is: How much Cu remains in solution? To answer the question, it is <em>assumed that none remains in solution</em>...(!) This leads to a slight correction (from <span class="math-container">$3\cdot 10^{-21}$</span> % to <span class="math-container">$4\cdot 10^{-21}$</span> %) Cu remaining. The text does ignore the small change in [S] that accompanies the change in [H<span class="math-container">$^+$</span>].</p> <blockquote> <p><em>If</em> there is a situation in which this very small difference matters, what is the right way to find it? </p> </blockquote> <p>My guess is that there would be a simple ODE governing this problem--but I'm not sure. </p> <p>The details above are included for completeness but the answer I think does not depend on the numbers here. </p> <p>*The source of this problem is Shaum's <em>College Chem.</em> (10 ed.), 18.13-14. </p> https://chemistry.stackexchange.com/q/102651 -1 How to distinguish between iron and nickel based on the mass of precipitate formed with sodium hydroxide? alina szoke https://chemistry.stackexchange.com/users/68717 2018-10-07T14:46:30Z 2018-11-06T17:21:01Z <blockquote> <p>For the reactions shown below, we added <span class="math-container">$\pu{5.00 mL}$</span> of <span class="math-container">$\pu{0.0390 M}$</span> <span class="math-container">$\ce{NaOH}$</span> to a test tube containing one of the two cations <span class="math-container">$\ce{Ni^2+}$</span> or <span class="math-container">$\ce{Fe^3+}$</span> and recovered <span class="math-container">$\pu{0.00695 g}$</span> of precipitate. <span class="math-container">\begin{align} \ce{Ni(NO3)2(aq) + 2 NaOH(aq) &amp;-&gt; Ni(OH)2(s) + 2 NaNO3(aq)}\\ \ce{Fe(NO3)3(aq) + 3 NaOH(aq) &amp;-&gt; Fe(OH)3(s) + 3 NaNO3(aq)} \end{align}</span></p> <ol> <li>How much precipitate in moles would be recovered theoretically if the ion was <span class="math-container">$\ce{Ni^2+}$</span>?</li> </ol> </blockquote> <p>For this answer I got <span class="math-container">$\pu{0.0000975 mol}$</span>, but that is wrong and I don't understand why.</p> <blockquote> <ol start="2"> <li><p>How much precipitate in moles would be recovered theoretically if the ion was <span class="math-container">$\ce{Fe^3+}$</span>? </p></li> <li><p>How much precipitate in grams would be recovered theoretically if the ion was <span class="math-container">$\ce{Ni^2+}$</span>?</p></li> <li><p>How much precipitate in grams would be recovered theoretically if the ion was <span class="math-container">$\ce{Fe^3+}$</span>?</p></li> </ol> </blockquote> https://chemistry.stackexchange.com/q/102034 1 Precipitation titration with Volhard method Anouk Lambrechts https://chemistry.stackexchange.com/users/68210 2018-09-22T18:11:26Z 2018-09-22T18:33:38Z <p>In the Vollhard method Chloride is precipitated from a solution using escess silver ions. <span class="math-container">$$\ce{Ag+_{(aq)} + Cl-_{(aq)} -&gt; AgCl_{(s)}}$$</span> After precipitation the solution is filtered and ammonium thiocyanate is added to the silver supernate with a small amount of ferric ammonium sulfate as an indicator where thiocyanate reacts preferentially with the silver:</p> <p><span class="math-container">$$\ce{Ag+_{(aq)} + SCN-_{(aq)} -&gt; AgSCN_{(aq)}}$$</span></p> <p>Once the all of the silver has reacted with thiocyanate the <span class="math-container">$\ce{Fe^3+}$</span> will then react to form <span class="math-container">$\ce{Fe[SCN]^2+}$</span> which produces a dark red color and indicates an equivalent point. and the chloride concentration is calculated by difference.</p> <p><span class="math-container">$$\ce{Fe^3+_{(aq)} + SCN-_{(aq)} -&gt; Fe[SCN]^2+_{(aq)}}$$</span></p> <p>Since the silver in the precipitate, why will you have to filter <span class="math-container">$\ce{AgCl}$</span> before you can start back titration instead of treating the solution? Would filtering also necessary for <span class="math-container">$\ce{AgBr}$</span> and <span class="math-container">$\ce{AgI}$</span>?</p> https://chemistry.stackexchange.com/q/98666 4 Mathematical explanation of blank titration in Mohr method for argentometry Mirko https://chemistry.stackexchange.com/users/33937 2018-06-23T16:33:22Z 2018-06-23T17:20:43Z <p>In my book it is written that in order to minimize systematic error in Mohr method for argentometry it is a common procedure to conduct a blank titration with $\mathrm{Ag^+}$ against a $\mathrm{CrO_{4}^{2-}}$ solution (with the same concentration of the original solution).</p> <p>The book then suggests to subtract the $\mathrm{Ag^+}$ volume needed to observe precipitation (in the blank solution) from the volume obtained as a result of the 'standard' titration.</p> <p>I don't understand why I have to subtract that number, specifically I do understand that there is a systematic error but I don't get why it is the volume of $\mathrm{Ag^+}$ obtained by the blank titration.</p> https://chemistry.stackexchange.com/q/96624 -1 Why do aluminum, magnesium, and calcium form white precipitate when mix with sodium hydroxide? user28603 https://chemistry.stackexchange.com/users/56854 2018-05-09T08:35:42Z 2018-05-09T10:49:25Z <p>Why do aluminum, magnesium, and calcium form white precipitate when mix with sodium hydroxide? Shouldn't there be no reaction? Because sodium is more reactive than all of those element, it should be impossible for sodium to be displace by those three elements.</p> <p><a href="https://www.youtube.com/watch?v=towhg_-GFrQ" rel="nofollow noreferrer">YouTube video - "NaOH + Al - sodium hydroxide and aluminum"</a></p> https://chemistry.stackexchange.com/q/95767 1 Precipitation of CaCl2 and FeCl3 in Na2CO3 [closed] Nehal Samee https://chemistry.stackexchange.com/users/58029 2018-04-23T18:02:36Z 2018-08-11T17:00:24Z <p>Why does $\ce{CaCl2}$ gives precipitation of $\ce{CaCO3}$ when $\ce{Na2CO3}$ is added, but $\ce{FeCl3}$ does not give the precipitation of $\ce{Fe2(CO3)3}$ in $\ce{Na_2CO_3}$ ?</p> https://chemistry.stackexchange.com/q/94897 -2 Amonium perchlorate [closed] Bruno Mondelo https://chemistry.stackexchange.com/users/62551 2018-04-09T08:16:02Z 2018-04-09T15:00:09Z <p>I want to make <strong>amonium perchlorate</strong>, having amonium chlorate. If I have a solution of <strong>sodium chlorate</strong> and I add amonium chlorate, will precipitate into amonium perchlorate? Or it is with a solution of <strong>sodium perchlorate</strong>?</p> https://chemistry.stackexchange.com/q/91880 0 Volume of NaOH needed to precipitate copper(II) but not calcium Desperados https://chemistry.stackexchange.com/users/57940 2018-03-07T17:12:40Z 2018-10-06T15:01:53Z <p>I have <span class="math-container">$\ce{NaOH}: C = \pu{0.1M}, V_{\ce{NaOH}}$</span> that I want to progressively add in a solution (in water) of: <span class="math-container">\begin{align} \ce{Cu^{2+}}&amp;: \pu{0.01M}; &amp; K_\mathrm{sp} &amp;= 10^{-18.6} \\ \ce{Ca^{2+}}&amp;: \pu{0.01M}, &amp; K_\mathrm{sp} &amp;= 10^{-5.26} \end{align}</span> with initial volume: <span class="math-container">$V_\mathrm{init} = \pu{0.1L}$</span>. I want to add <span class="math-container">$\ce{NaOH}$</span> as long as <span class="math-container">$\ce{Cu(OH)2}$</span> precipitates, without causing any precipitation of <span class="math-container">$\ce{Ca(OH)2}$</span>.</p> <p>Knowing that the equilibrium will be displaced to the left and I will have precipitation, I have done: <span class="math-container">$$\begin{array}{ccc} \ce{Cu(OH)2 &amp;-&gt;&amp;Cu^2+ &amp;+ &amp;2OH^-}\\ &amp;&amp;0.001&amp;&amp;0.1V_{\ce{NaOH}}\\ &amp;&amp;0.001-x&amp;&amp;0.1V_{\ce{NaOH}} - 2x\\ \end{array}$$</span></p> <p>Also, since I want to achieve <span class="math-container">$99.999\%$</span> precipitation of <span class="math-container">$\ce{Cu(OH2)}$</span> I have got: <span class="math-container">$$x = 99.999\cdot10^{-5}$$</span></p> <p>Using all that in the expression of <span class="math-container">$K_\mathrm{sp}$</span> I get : <span class="math-container">$$10^{-18.6} = \left (\frac{0.001 - 99.999 \cdot 10^{-5}}{0.1+V_{\ce{NaOH}}} \right)\cdot\left(\frac{0.1\cdot V_{\ce{NaOH}} - 2\cdot99.999\cdot10^{-5}}{0.1 + V_{\ce{NaOH}}}\right)^2$$</span> </p> <p>However for this expression I get a value of <span class="math-container">$V_{\ce{NaOH}}$</span> that is extremely small. Is there anything wrong with my reasoning and if yes, how can I determine the volume for which i get the precipitation I want?</p> https://chemistry.stackexchange.com/q/91123 0 Is it absolutely necessary to use centrifuge to separate ethylene glycol from precipitate? Utkarsh Singh https://chemistry.stackexchange.com/users/59395 2018-02-22T10:16:22Z 2018-08-21T16:02:19Z <p>I've been trying to synthesize intermetallics of $\ce{XY}$ type using transition metal salts and main group element salts. For my system, EG, DEG, and TEG are good solvents. However, because of them being highly viscous, I've experienced normal drying doesn't do the trick and that the precipitate remains puffy even after drying (which shouldn't be the case) before sintering but by using centrifugation, I've been successful at removing the puffy texture of the precipitate. </p> <p>Recently, I came upon <a href="https://drive.google.com/open?id=1jal6q3_EtOj4d6k5Kxvudl4VpdixWFx3" rel="nofollow noreferrer">this</a> paper where they haven't used centrifugation for the main sample. Could anyone please explain how does this work? </p> https://chemistry.stackexchange.com/q/89847 0 Hydrolysis of transition metals' halides? tox123 https://chemistry.stackexchange.com/users/7786 2018-02-01T01:53:03Z 2018-02-01T15:20:34Z <p>So, it's a relatively common known solubility rule that any hydroxide with a cation not in the first two groups is basically insoluble. So supposing we have a transition or post-transition metal $M$ and say it forms a charge of $n+$ (so it's cation is $\ce{M^{n+}}$) why would $\ce{MX_n}$ (where $\ce{X}$ is a halide) remain in solution? Why doesn't the following reaction occur? $$\ce{MX_n(aq)} + n\ \ce{H_2O(l)} \rightarrow \ce{M(OH)_n(s)} + n\ \ce{HX(aq)}$$ My thought was that perhaps this reaction does occur, but just very slowly and not fast enough for the $\ce{M(OH)_n}$ to precipitate in noticeable amounts. However, I don't really know and this is just a guess. Could someone explain to me why this doesn't happen?</p> https://chemistry.stackexchange.com/q/89297 -1 Determine concentration of phosphate after copper(II) oxalate is precipitated Mabadai https://chemistry.stackexchange.com/users/55298 2018-01-22T17:06:13Z 2018-01-23T09:50:42Z <p>Solution with $\ce{Na3PO4}$ with $\pu{0.02 M}$ concentration and $\ce{Na2C2O4}$ with $\pu{0.03 M}$ concentration. I add excess $\ce{Cu(NO3)2}$ very carefully – tiny amounts. </p> <p>What will be my $\ce{PO4^3-}$ concentration at the moment when precipitation of $\ce{CuC2O4}$ will occur?</p> <p>Given: $K_\mathrm{sp}\left(\ce{Cu3(PO4)2}\right) = \pu{1.40e-37}$, $K_\mathrm{sp}\left(\ce{CuC2O4}\right) = \pu{4.43e-10}$.</p> <p>I know that the reactions will be:</p> <p>\begin{align} \ce{Cu3(PO4)2 &amp;&lt;=&gt; 3Cu^2+ + 2PO4^3-}\\ \ce{CuC2O4 &amp;&lt;=&gt; Cu^2+ + C2O4^2-} \end{align}</p> <p>Now, solving some math I find that $\ce{Cu3(PO4)2}$ precipitates first. But then I start asking my self what is happening with the $\ce{Cu}$ amount. Never did lab lessons or thought about the process behind the scene.</p> https://chemistry.stackexchange.com/q/88938 7 How much calcium hydroxide will precipitate after addition of sodium hydroxide into saturated calcium hydroxide solution? George Tian https://chemistry.stackexchange.com/users/47110 2018-01-16T05:17:30Z 2018-07-15T10:14:19Z <p>Below was question 34 in the USNCO 2017 exam:</p> <blockquote> <p>If $\pu{0.10 mol}$ of solid $\ce{NaOH}$ is added to $\pu{1.00 L}$ of a saturated solution of $\ce{Ca(OH)2}$ $(K_\mathrm{sp} = \pu{8.0 \times 10^-6})$, what percentage of the calcium hydroxide will precipitate at equilibrium?</p> <p>(A) Roughly 50%<br> (B) Roughly 75%<br> (C) Roughly 95%<br> (D) Over 99% </p> </blockquote> <p>My solution is as follows:</p> <ol> <li><p>Find concentration of $\ce{Ca^2+}$ $(\pu{0.02 M})$ and $\ce{OH-}$ $(\pu{0.04 M})$ ions from dissolved calcium hydroxide using $K_\mathrm{sp}$.</p></li> <li><p>Add hydroxide ion concentration from sodium hydroxide (assuming full dissolution) to get total hydroxide concentration of $\pu{0.14 M}$</p></li> <li><p>Find reaction quotient $Q = 0.02 \times 0.14^2 = 3.92 \times 10^{-4}$</p></li> <li><p>Find amount of calcium $(x)$ and hydroxide ions $(2x)$ that will precipitate at equilibrium by using algebraic equation: $$(0.02 - x)(0.14 - 2x)^2 = 8.0 \times 10^{-6}, x = \pu{0.019 M}$$</p></li> <li><p>Find percentage of calcium hydroxide precipitated: $$\frac{0.019}{0.02} \times 100\% = 95\%,$$ hence (C)</p></li> </ol> <p>I am unsure about step 4, where a cubic equation appears, and would not be able to be solved in exam conditions (use of graphing calculator is not permitted).</p> <p>Is there a simpler method?</p> https://chemistry.stackexchange.com/q/88514 11 Why exactly does precipitation occur? Shikhar_Mohan https://chemistry.stackexchange.com/users/57315 2018-01-08T09:51:03Z 2018-02-02T21:01:31Z <p>In a solution, we have ions floating around but when we have a precipitate, they're arranged as they would be in a solid. This conversion should cause a decrement in the entropy of the system and cause the reaction to be non-spontaneous, but that is not the case. Data of several $K_\mathrm{sp}$'s suggests that the reactions are very spontaneous, with $\log{K_\mathrm{eq}}$ around $30-40$ for some salts. Why does this happen?</p> <p>One possible explanation I could think of was that all precipitation processes are enthalpy driven instead of entropy driven but in those cases enthalpy should either be really high and all solubilities must increase upon increase in temperature. I haven't studied these enough to know if that is the case or not, so help from someone more experienced would be appreciated.</p>