Question: $\ce{H2S}$ ($\pu{0.1M}$; $K_\mathrm{a} = 1.2×10^{-20} $) and $\ce{HCl}$ ($\pu{0.3M}$) with same volume are mixed together. What is the resultant $\mathrm{pH}$?

My approach: Let the volume of each be $x \; \pu{L}$ $$ \begin{array}{lcc} &\ce{H2S & -> & 2H+ & + & S^2-} \\ \text{initially} & 0.1 && 0 && 0 \\ \text{after} & 0.1(1-\alpha) && 2 \times 0.1\alpha && 0.1\alpha \end{array} $$ $$ \begin{array}{lcc} &\ce{HCl & -> & H+ & + & Cl-} \\ \text{initially} & 0.3 && 0 && 0 \\ \text{after} & 0 && 0.3 && 0.3 \end{array} $$ $$K_\mathrm{a} =\frac{\ce{[H]}^2\ce{[S^2-]}}{\ce{[H2S]}}$$ $\ce{[H]} = $ total moles of $\ce{H}$/Total volume $=\frac{0.2\times x\times\alpha+0.3x}{2x}=\frac{0.2\times\alpha+0.3}{2}$

$0.2×\alpha$ can be neglected and $(1-\alpha)$ can be taken as $1$. $$K_\mathrm{a} = \left[\frac{0.3}{2}\right]^2[\alpha]$$ Calculating for $\alpha$, we get it as $5.33\times 10^{-19} $ And we get $\ce{[H]}$ as $0.15$.

$$\therefore \mathrm{pH} =0.8$$ Is my approach wrong???

Another approach: I have been told by my teacher that resultant $\mathrm{pH}$ will nearly be the $\mathrm{pH}$ of strong Acid but if we add Weak acid will it not dilute the strong acid. Here in this case make the volume twice.


closed as off-topic by Todd Minehardt, Mithoron, M.A.R. ಠ_ಠ, Buck Thorn, Nuclear Chemist Jun 1 at 6:51

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    $\begingroup$ There is nothing "another" about it. You've just described the same approach twice. $\endgroup$ – Ivan Neretin May 15 at 17:47
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    $\begingroup$ You're getting HCl with two times lower conc. - H2S is irrelevant. BTW first Ka of H2S is much higher then what you wrote but still not enough to make a difference here. $\endgroup$ – Mithoron May 15 at 17:50
  • $\begingroup$ But adding H2S will not change the concentration of the solution?? $\endgroup$ – Satwik May 15 at 17:59
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    $\begingroup$ You aren't adding just H2S. You're adding H2S and a good deal of water. Of course water will dilute the solution. $\endgroup$ – Ivan Neretin May 15 at 18:05
  • $\begingroup$ So it should change the concentration of the solution. So will the pH will not be affected as the solution becomes dilute on adding h2s solution. So do we have to consider this dilution while calculationg the pH??? If yes then if we are calculating pH of hcl only and neglecting of h2s as it will be very small then also the concentration of solution will change altering the pH $\endgroup$ – Satwik May 15 at 18:20

Question: $\ce{H2S}$ ($\pu{0.1M}$; $K_\mathrm{a} = 1.2×10^{-20} $) and $\ce{HCl}$ ($\pu{0.3M}$) with same volume are mixed together. What is the resultant $\mathrm{pH}$?

Since, $\ce{HCl}$ is strong acid, we can assume all $\ce{HCl}$ molecules are dissociated in aqueous solution to begin with. Let the volume of each solution be $\pu{1.0 L}$ and $\ce{H2S}$ is dissociated $\alpha$ amount in aqueous solution at the equilibrium after combined. $$ \begin{array}{lcc} &\ce{HCl + H2O & -> & H3O+ & + & Cl-} \\ \text{initial amounts} & - && \pu{0.3mol} && \pu{0.3mol} \\ &\ce{H2S + H2O & -> & 2H3O+ & + & S^2-} \\ \text{initial amount} & \pu{0.1mol} \;(\ce{H2S}) && \pu{0.3mol} && 0 \\ \text{amount at equilibrium} & \pu{(0.1-\alpha) mol} && \pu{(0.3+ 2\alpha) mol} && \pu{\alpha \; mol}\\ \text{conc. at equilibrium} & \frac{\pu{(0.1-\alpha) mol}}{\pu{2.0 L}} && \frac{\pu{(0.3+ 2\alpha) mol}}{\pu{2.0 L}} && \frac{\pu{\alpha \; mol}}{\pu{2.0 L}} \end{array} $$ Now, we can define $K_\mathrm{a}$ in terms of $\alpha$:

$$K_\mathrm{a} =\frac{\ce{[H]}^2\ce{[S^2-]}}{\ce{[H2S]}} = \frac{\left(\frac{\pu{(0.3+ 2\alpha) mol}}{\pu{2.0 L}}\right)^2\left(\frac{\pu{\alpha \; mol}}{\pu{2.0 L}} \right)}{\left(\frac{\pu{(0.1-\alpha) mol}}{\pu{2.0 L}}\right)} = \pu{\frac{\alpha ( 0.3+ 2\alpha)^2}{4.0(0.1-\alpha)} mol^2L^{-2}}$$

Since $K_\mathrm{a} = 1.2×10^{-20} $, we can assume $\alpha \lt\lt\lt 0.1 $, and hence,

$$K_\mathrm{a} \approx \pu{\frac{\alpha}{0.1}\left(\frac{0.3}{2.0}\right)^2 mol^2L^{-2}}$$ When resolve for $\alpha$, we got: $\alpha = \pu{5.33\times 10^{-20} mol\:L^{-1}}$, and hence, at equilibrium, $\ce{[H]} \approx \pu{\frac{0.3}{2} mol\:L^{-1}} \approx \pu{0.15 mol\:L^{-1}}$.

$$\therefore \mathrm{pH} =0.82$$

Although you got the anticipated answer, your setup is not correct, e.g., see after concentrations (or amounts) of $\ce{H2S}$ dissociation amounts at equilibrium.

  • $\begingroup$ :$K_{a1}=10^{-7},K_{a2}=10^{-19}$,How$ K_a=1.2\times{10^{-20}}$ ? $\endgroup$ – Adnan AL-Amleh May 18 at 2:21

Paraphrasing from the excellent comments (even if the commenters might have been exasperated):

The pH of a mixture of $\ce{H2S}$ solution with $\ce{HCl}$ solution is about the same as the corresponding mixture of water with $\ce{HCl}$ solution. The reason is that $\ce{H2S}$ does not significantly dissociate in water, and even less so in an acidic solution (difference between pKa and pH on the order of 19). So adding $\ce{H2S}$ solution lowers the concentration of $\ce{H+}$, just like adding pure water.

In your answer, you are leaving out the units and don't say what quantities you are showing (amounts or concentrations). If it is concentrations, the concentrations of the solutions before and after mixing are different, and you are not accounting for it. For example, the total concentration (dissociated or not) of HCl is 0.3 mol/L before mixing, and 0.15 mol/L after.


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