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1

Based on my comment above, this is what I think the table should look like. \begin{array}{c | c c c c c c} \text{RXN} &\ce{Zn^2+}\text{(aq)} & \ce{CH3COO-}\text{(aq)} & \ce{Na+OH-}\text{(aq)} & \ce{Zn(OH)2}\mathrm{(s)} \\\hline \text{I} &\pu{0.05 M} & \pu{0.10 M} &\pu{ 0.025 M} &\pu{0 moles} \\ \text{...


0

Taking into my consideration to the above comment, I solved the problem. I used the following table to determine the amount of each species at the end of the reaction: \begin{array}{c | c c c c c c} \text{RXN} &\ce{Zn(CH3COOH)2}_\text{(aq)} & + &\ce{2NaOH}_\text{(aq)} &\ce{->}& \ce{2CH3COONa}_\text{(aq)} &+& \...


0

This is a simplistic question which doesn't require a great deal of analysis. When ionic salts dissolve the typical assumption is that they completely ionize. So: $$\ce{Ca(OH)2 -> Ca^2+ + 2OH-}$$ $$\ce{Mg(OH)2 -> Mg^2+ + 2OH-}$$ The assumption is that the calcium hydroxide will dissolve completely. However the magnesium hydroxide is less soluble and ...


3

Strontium hydroxide is a strong base, so you can calculate $[\ce{OH-}]$ as $\pu{0.02 M}$, then use $$K_\mathrm{w} = \ce{[OH-][H3O+]}$$ $$1\cdot 10^{-14} = 0.02\cdot [\ce{H3O+}] \quad\to\quad [\ce{H3O+}] = \pu{5e-13 M} \quad\to\quad \mathrm{pH} = 12.3$$


3

The solubility of calcium salts is highly dependent on pH. For example, let’s look at tricalcium phosphate, which presents the solubility equilibrium: $$\ce{Ca3(PO4)2(s) + H2O <=> 3Ca^2+(aq) + 2PO4^3-(aq)}$$ The solubility product constant for this equilibrium ($K_\mathrm{sp}$) is vastly varied from source to source. Some listed it as $\pu{2.0 × 10^–29 ...


0

It is because the phosphate ion $PO_4^{-3}$ is a weak base. In general, ionic substance with anions that are weak bases dissolve better in acidic solution.


2

Pure water (rain as well as distilled water) in equilibrium with the atmosphere ($p_{\ce{O2}}=10^{-3.5}\ \mathrm{atm}$) can be calculated to contain about $$\begin{align} \mathrm{pH}=-\log[\ce{H+}]&=5.65\\ -\log[\ce{HCO3-}]&=5.65\\ -\log[\ce{CO3^2-}]&=10.3\\ -\log[\ce{H2CO3^*}]&=5.0\\ -\log[\ce{CO2}]&=5.0\\ -\log[\ce{H2CO3}]&=7.8\\ \...


0

Aside of influence on the crystallization rate, you have also consider the direct chemical influence. The sodium carbonate hydrolyzes in not enough alkaline solutions. $$\ce{Na2CO3 + H2O <=> NaHCO3 + NaOH}$$ Precipitation of carbonate is interfered by solubility of Bicarbonates. $$\ce{CaCO3 v + H2O + HCO3- <=> Ca(HCO3)2 + OH-} $$ Addition of ...


1

At low pHs there is little "free" $\ce{CO3^{2-}}$ to precipitate the $\ce{Mg^{2+}}$ and $\ce{Ca^{2+}}$ cations. Most of the carbonate species are dissolved $\ce{CO2}$, $\ce{H2CO3}$ and $\ce{HCO3-}$. Thus the precipitate forms slowly and you get relatively large crystals. At high pHs there is a lot "free" $\ce{CO3^{2-}}$ to precipitate the $\ce{Mg^{2+}}$ ...


1

Any good general chemistry textbook has a chapter on electrolysis. If you wish to add more scholarly work check the Journal of Chemical Education. https://pubs.acs.org/action/doSearch?text1=Electrolysis&quickLinkYear=&quickLinkVolume=&field1=Title&type=within&publication=346464552 In the first part you electrolyze a solution of salt. ...


1

Edit: At the equivalence point, the solution contains the dissolved salt $\ce{CH3NH3Cl}$, dissociated to $\ce{CH3NH3+ + Cl-}$, as @MaxW noted. You can calculate $pH$ of the conjugate acid $\ce{CH3NH3+}$ at its dissociation equilibrium. $$\begin{align} K_w &= K_a . K_b = 10^{-14}\\ pK_w &= pK_a + pK_b = 14\\ \ce{ CH3NH3+ &<=> H+ + CH3NH2 }...


0

The following link from a very old text (pages 279-280) suggests boiling the wax in an alkaline solution. This method may be specific to that wax described as a new vegatable wax from a laurel like shrub from the forests of Para and Bahia. Please explain the application. Why do you want the wax to be alkaline? What are you going to do with the wax?


3

Although pH scale is rather strictly used for aqueous solutions, the concept of pH in organic or non-aqueous solutions is not trivial. It is called apparent pH or operational pH when organic solvents are present. There is a practical problem here as well, do you think your buffers will dissolve in ethanol? Once you figure this out, you can read a section in ...


2

To solve this I used the Henderson-Hasselbalch equation: $$\ce{pH} = \mathrm{p}K_\mathrm{a} + \log \left(\frac{[\ce{A^-}]}{[\ce{HA}]}\right)$$ Because $\ce{pH} = -\log(\mathrm{p}K_\mathrm{a})$ we end up with: $\ce{pH} = -\log(\mathrm{p}K_\mathrm{a})+ \log \left(\frac{1}{3}\right)=-\log \left(6.7 \times 10^{-5}\right) + \log \left(\frac{1}{3}\right)= 3....


2

In a chemical sense, "anode" of an electrolytic cell is the most powerful oxidizing reagent known, so much so that it oxidize F(-) to elemental fluorine. Don't forget that the electron balance in an electrolytic cell is always maintained so the gram-equivalents of x reduced at the cathode must equal gram equivalents of y oxidized at the anode. As a result, ...


0

Any aqueous solution increases its buffer capacity toward $\mathrm{p}H = 0$ or $\mathrm{p}H = 14$, as more and more acid or basis is needed to change $\mathrm{p}H$ by a given interval. It happens even without presence of any buffering substance, aside of strong acids or bases. If an acid, monoprotic or diprotic, is added, a peak(s) of maximum buffer ...


1

Can I multiply Ka1 and Ka1 to eliminate [$\ce{C4H5O6−}$], and then get the concentration of C4H4O62− necessary by plugging in 0.1 M for [C4H6O6] and the target pH in the appropriate form in [H+] No, because $\ce{C4H5O6−}$ is a one of the major species. In fact, if you add the tartaric acid and its double salt at equimolar ratios, $\ce{C4H5O6−}$ will be the ...


0

I would take different approach. From given $pH$ and acidity constants, calculate ratios of the particular forms of the tartaric acid. From given total molar amount, you get the molar amounts for the particular forms. From molar amount of particular forms, you can get molar amount of a hydroxide to be added to the tartaric acid, or of a strong acid to be ...


0

Part A - Calculating the initial $\pu{pH}$(before adding $\pu{0.02 mol}$ of $\ce{HCl})$ The equilibrium here is $$\ce{CH3CH2NH2 + H2O <=> CH3CH2NH3+ + OH−}$$ For $\ce{ CH3CH2NH2} , \mathrm{K_b} = 4.6\times{10^{−4} }$ $$\mathrm{K_b} = 4.6\times{10^{−4} }=\frac{[\ce{CH3CH2NH3+}][\ce{OH−}]}{ [\ce{CH3CH2NH2}]}$$ The nominal concentrations of the ...


1

Generally, the maximum buffer capacity is at $\ce{pK_a}$ . The tartaric acid is somewhat special for 2 reasons: It is a diprotic acid with both $\ce{pK_a}$ very close, with the $\ce{pK_{a1}}$ rather low, being affected by the reason 2. : $$\ce{pK_{a1}}=2.89,\ce{pK_{a2}}= 4.40 (L+)$$ The solution buffer capacity (not limited to presence of specific buffer ...


2

If we consider $\ce{HA}$ as a weak acid, then at the half equivalence point, $$\mathrm{p}H = \mathrm{p}K_\mathrm{a}$$ As $$\mathrm{p}H = \mathrm{p}K_\mathrm{a} + \mathrm{p[\ce{HA}]} -\mathrm{p[\ce{A-}]}$$ and for the half equivalence point, $$\mathrm{p[\ce{HA}]} =\mathrm{p[\ce{A-}]}$$ So the higher the $\mathrm{p}K_\mathrm{a}$ is, the higher is $\mathrm{p}H$ ...


0

$K_a = \dfrac{\ce{[H^+][A^-]}}{\ce{[HA]}}$ Let's assume a weak acid so that we'll assume a negligible fraction ionizes. So for all concentrations we get $\ce{[H^+] = [A^-]} = \sqrt{(K_a)\ce{([HA])}}$ and the dissociation degree (fraction) is: $\dfrac{\ce{[A^-]}}{\ce{[HA]}} = \sqrt{\dfrac{K_a}{\ce{[HA]}}}$ So if the concentration of HA is reduced by a ...


1

You are on the right track according to the directions given to you ( = the problem statement), but the directions themselves are wrong. Say you had an equilibrium with $\ce{{[H^+][A^-]\over[HA]}=K}$. You dilute it five-fold and tell your particles to wait a bit with dissociation until you are done with the math. That makes all concentrations 5 times ...


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