New answers tagged

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Instead of oxygen bubbling, add some hydrogen peroxide $\ce{H2O2}$. It is not expensive. And it works perfectly well. Here is a formula that does your job in less than 1 minute, without any foam. Dissolve $\ce{20 g CuCl2·6 H2O}$ in $\pu{100 mL}$ water. Add $\ce{100 mL HCl}$ $35$% and $\ce{100 mL H2O2}$ $30$%. Here you are. Dip your circuit board in this ...


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The circuit etch solution is made from CuCl$_2$ rather than CuCl. https://www.instructables.com/PCB-Etching-Solution-Cupric-Chloride/ After etching/dissolving some copper, you would get CuCl, and regeneration involves reoxidizing the Cu$^+$ to Cu$^{2+}$. When you use air, not all the O$_2$ is captured, and none of the N$_2$, so you have lots of gas bubbles ...


2

So you need $$K_p = \frac{p_{\ce{H2S}}^3}{p_{\ce{H2}}^3}$$ where $p_{\ce{H2S}}$ and $p_{\ce{H2}}$ are partial pressures of $\ce{H2S}$ and $\ce{H2}$ respectively. Partial pressure is nothing but the mole fraction of the substance times the total pressure. You already have mole fractions, so that part is done. Now notice that in the expression of $K_p$ , you ...


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You didn't indicate whether you want $\%(w/w)$ or $\%(w/v)$ in your post. If you want $\%(w/v)$, you can easily calculate that by given data: $$\pu{6 M} \ \ce{HCl} = \frac{\pu{6 mol} \ \ce{HCl}}{\pu{1 L} \ \text{solution}} = \frac{\pu{6 mol}\times \frac{\pu{36.46 g}}{\pu{1 mol}} \ \ce{HCl}}{\pu{1 L} \ \text{solution}} = \frac{\pu{218.76 g} \ \ce{HCl}}{\pu{1 ...


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This question was answered well enough 5-7 years ago, but there is something still lacking. The resonance diagrams drawn in brackets suggest that resonance is responsible for increasing the acidity of phenol; I find the wording misleading. Better would be that the resonance of the phenyl ring is responsible for the stabilization of phenolate anion. IMHO, ...


3

The proton transfers may happen in cascade. For example, pure $\ce{HCl}$ is a gas that dissolves and reacts with water (acting as a base) according to a proton transfer from HCl to $\ce{H2O}$ : $$\ce{HCl + H2O -> H3O+ + Cl- ... \ (1)}$$ But $\ce{H2O}$ does not hold its proton very strongly. Here is why : if another base like ammonia is added to this ...


1

The PH at equivalence point of titration for a strong acid and strong base should be 7 theoretically specking because, they produce a neutral salt on reaction. This means that the salt wont be hydrolysed or anything like that to raise the PH above 7. But during actual titration its important that we use an indicator to map the equivalence point. Generally ...


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The equivalence point should be at pH 7, if it is higher it means that too much NaOH was added. Seems like you should indeed add more points. https://www.khanacademy.org/test-prep/mcat/chemical-processes/titrations-and-solubility-equilibria/a/acid-base-titration-curves


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Let's say you are titrating a weak acid $\ce{H3A}$ with a strong base. $$\ce{H3A + OH- -> H2A- + H2O}$$ $$\ce{H2A- + OH- -> HA^{2-} + H2O}$$ $$\ce{HA^{2-} + OH- -> A^{3-} + H2O}$$ When all $\ce{H3A}$ is reacted with $\ce{OH-}$ to form $\ce{H2A-}$, the first equivalence point is reached. Before the first equivalence point is reached, the $\ce{H3A - ...


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Well ! Titration is an operation that has to be done with the highest precision. The necessary end-point volume has to be determined at ± 1 drop of reactant, using an apparatus (pH meter, conductimeter, colorimeter, etc.) or a colored indicator, changing color at the end-point. Such a titration can be done with an acid-base reaction. But it can be done with ...


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No. While neutralization is often used in titration, other kinds of reactions may also be adapted to this laboratory technique. Precipitation for the measurement of chloride ion concentration is a well-known example.


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Agar-agar consists of polysaccharides; Wikipedia for example mentions agarose: (credit) As with many naturally occurring polymers, the degree of polymerization (i.e., the repeating number of building blocks per polymer chain) varies. Any form of processing may increase the variation with shorter and longer chains of this polymer. The hydroxyl groups may ...


1

The other answer has given the major conformer of the protonated form of the o-hydroxybenzoic acid, but I believe there is a part of your question that still remains to be answered. You asked why the hydrogen bond makes the hydrogen bond more acidic than the para isomer, when the hydrogen bond is already present in the protonated form of the ortho-isomer. ...


3

Taking the data from this answer by ron, the equilibrium constant for the keto-enol tautomerism seen in acetone is $$K_\mathrm{eq} = \frac{\text{[enol]}}{\text{[carbonyl]}}$$ $$ \begin{array}{lc} \hline \text{compound} & K_\mathrm{eq} \\ \hline \text{acetaldehyde} & 6 \times 10^{-7} \\ \text{acetone} & 5 \times 10^{-9} \\ \hline \end{array} $$ ...


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$6.6$ is neutral for pretty much all purposes. So is $6.0$. So is $5.0$. You need to go way lower than that. Table vinegar may have $\rm{pH}\approx2.5\dots3.0$, and that's what you want to use to hear that gentle fizz of dissolving carbonates. Yes, this is a viable method. No, it will not corrode steel pipes to any significant extent, provided that you apply ...


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Reducing the pH of tap water to $6.6$ or $6.9$ will not corrode pieces of Zinc or Iron metals. Even pH $5$ is not sufficient.


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Acid/base nature of any reagent is a relative term. For instance, water $(\mathrm{p}K_\mathrm{w} = 14)$ also acts as an acid or base based on which reagent it is reacting with. Basically, any reagent with $\mathrm{p}K_\mathrm{a} \lt 7$, water acts as an base (e.g., acetic acid; $\mathrm{p}K_\mathrm{a} = 4.75$): $$\ce{CH3COOH + H2O <=> CH3COO- + H3O+}$$ ...


-1

please check out our paper we published already in 2008: Chances and pitfalls of chemical cross-linking with amine-reactive N-hydroxysuccinimide esters. Kalkhof S, Sinz A. Anal Bioanal Chem. 2008 Sep;392(1-2):305-12. doi: 10.1007/s00216-008-2231-5 Another paper on NHS ester reactivity is this: Stefanie Mädler, Claudia Bich, David Touboul, Renato Zenobi, ...


2

Some of the experimental details are missing, but if you recall $$\mathrm{Abs} = \varepsilon \cdot c \cdot d$$ about the absorption depending on the molar extinction coefficient $\varepsilon$, the analyte concentration $c$, and the optical path length $d$, it could be due to a low concentration of the analyte. Independent of the former, given the $\pi$ ...


0

Ammonium hydroxide does not exist. $\ce{NH4OH}$ is a formula that was invented in the 19th century to explain why $\ce{NH3}$ is so easily and exothermically dissolved into water, as if there was a chemical reaction with water. But it has been proven that as soon as an ion like $\ce{NH4^+}$ meets the ion $\ce{OH^-}$, the following reaction occurs : $$\ce{NH4^+...


9

The correct order is in fact $(\bf{d}) \gt (\bf{b}) \gt (\bf{c}) \gt (\bf{a})$, the reason for this is as follows. $(\bf{a})$ is definitely last in this order of basicity since its lone pair is delocalised by the phenyl ring. Now we need to compare the other three. We can split this into two parts, a comparison between $(\bf{d})$ and $(\bf{b})$ $(\bf{b})$ ...


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Some of the conformers of salicylic acid are given below, all of which are planar in configuration.$\mathrm{^{[1]}}$ Although the compound can adopt several conformations the first conformer(in the first image, the structure on the left) is of the lowest energy and has a strong intramolecular hydrogen bond. The second conformer(first image, right structure) ...


4

According to Bronsted-Lowry Acid-Base theory, stronger acids have weaker conjugate bases and vice-versa. $\ce{HCl}$, being a strong acid in water, it's conjugate base i.e $\ce{Cl-}$ is weak or it is more stable. Although it reacts with $\ce{H2O}$ but to a very less extent. Whereas in the case of $\ce{CH3COOH}$, it being a weak acid, it's conjugate base is a ...


3

Do you say stainless steel? Stainless steel is an alloy of $\ce{Ni, Cr, Fe}$ with other trace elements, and owes its apparent resistance to corrosion to a protective, adherent, coating of mixed chrome, nickel, and iron oxides. A large amount is probably $\ce{Cr^3+}$, which is amphoteric and will dissolve in hydroxide solution. Once the protective coating is ...


3

The reaction of alcohol + NaOH producing $\ce{C2H5ONa}$ + water is an equilibrium which is usually strongly pushed to the left hand side. But in the presence of $\ce{CS2}$, the alkoxide $\ce{C2H5ONa}$ is transformed into xanthate. So the equilibrium is driven to the right hand side.


3

If a solution of $\ce{H2SO4}$ is added to a some carbonate, bicarbonate or sulfite, it will produce some $\ce{CO2}$ or $\ce{SO2}$ effervescence with gas bubbles filling the container where this test is done, and maybe cause it to overflow. With sodium bicarbonate the equation is : $$\ce{H2SO4 + 2 NaHCO3 -> Na2SO4 + 2 H2O + 2 CO2}$$ With sodium sulfite, ...


0

AFAIK, it is $\ce{[H+]}=\sqrt{K_\mathrm{a1} \cdot c_1 + K_\mathrm{a2} \cdot c_2}$, with assumption the respective $c \gg [\mathrm{H+}] \gg [\mathrm{OH-}]$


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Reused my answer to how-to-numerically-model-a-phosphoric-acid-titration-curve, edited for diprotic cases: First, calculate the common denominator $CD$: $$a_1 = [\ce{H+}]^2$$ $$a_2 = [\ce{H+}] \cdot K_\mathrm{a1} = a_1 \cdot \frac {K_\mathrm{a1}}{[\ce{H+}]}$$ $$a_3 = K_\mathrm{a1} \cdot K_\mathrm{a2} = a_2 \cdot \frac {K_\mathrm{a2}}{[\ce{H+}]}$$ $$CD = a_1 +...


0

Let's admit that some $\ce{H2S}$ is added into a solution containing both ions $\ce{Zn^{2+}}$ and $\ce{Fe^{2+}}$ with a concentration $\pu{0.1 M}$. In my table, the solubility products of $\ce{FeS}$ is $\pu{6.3 10^{-18}}$, which is $10$ times higher than your value : $\pu{6.3 10^{-19}}$. With your value for $\ce{ZnS}$ and mine for $\ce{FeS}$, the maximum ...


0

When you acidify a solution, you increase the concentration of hydronium ions and while doing so decrease the concentration of hydroxide ions. This decrease in the concentration of hydroxide ions cause the equilibrium of the hydrolysis(both complete and partial) of the sulphide ion to shift forward. $$\ce{S^2- + 2H2O <=> H2S + 2OH-}$$ This forward ...


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