pH of strong acid weak acid mixture [closed]

Question

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.

Answer 1

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.

Answer 2

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.