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I am trying to calculate the $\mathrm{pH}$ of a solution obtained by mixing $\pu{20.00 mL}$ of $\pu{0.2 M}$ potassium hydrogen phosphate solution and $\pu{10.00 mL}$ of $\pu{0.25 M}$ hydrochloric acid solution $(K_{a2} = \pu{6.34 \times 10^{-8} mol/L})$.

My solution:
So, hydrogen phosphate reacts with oxonium to form dihydrogen phosphate: $$\ce{HPO4^2- + H3O+ <=> H2PO4- + H2O}$$ The equilibrium concentrations of hydrogen phosphate and dihydrogen phosphate are: $$[\ce{HPO4^2-}] = c_0(\ce{HPO4^2-}) - [\ce{H3O+}] \quad \text{and} \quad [\ce{H2PO4-}]=[\ce{H3O+}]$$ where $[\ce{H3O+}] = \frac{V(\ce{HCl})}{V(\ce{HCl)} + V(\ce{HPO4^2-)}}c_0(\ce{HCl})$.

Finally, I plugged everything in the Henderson–Hasselbalch equation $$\text{pH} = \text{p}K_\text{a2} + \log{\frac{[\ce{HPO4^2-}]}{[\ce{H2PO4-}]}}.$$

I got $\mathrm{pH}$ of 7.34, but the answer sheet says 6.97. What did I do wrong?

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    $\begingroup$ What concentrations did you calculate? You have a good explanation of the correct answer from Poutnik, but I'm not sure it is clear how you got 7.34, and where the error was. $\endgroup$
    – Karsten
    Commented Nov 7, 2023 at 17:31

2 Answers 2

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There is initially $\pu{0.02 L} \cdot \pu{0.2 mol/L} = \pu{0.004 mol}\ \ce{HPO4^2-}$
There is added $\pu{0.01 L} \cdot \pu{0.25 mol/L} = \pu{0.0025 mol}\ \ce{H+}$

This leads to approximately $\pu{0.0025 mol}\ \ce{H2PO4-}$ and $\pu{0.0015 mol}\ \ce{HPO4^2-}$.

As $\text{p}K_\text{a} = -\log {(\pu{6.34E-8})} \approx 7.20$,

then $\text{pH} \approx \text{p}K_\text{a} + \log {\left(\frac{n(\ce{HPO4^2-})}{n(\ce{H2PO4-})}\right)}=7.20 + \log{\frac{0.0015}{0.0025}} = 6.976$

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  • $\begingroup$ @Poutnik. Sorry ! One zero is missing in the calculation of the ions after mixing. Mixing $4$ millimol $\ce{HPO4^{2-}}$ + $2.5$ millimol $\ce{H+}$ does not produce $25$ millimol $\ce{H2PO4^-}$ and $15$ millimol $\ce{HPO4^-}$ as written, but it produces $10$ times less. This mixture produces $2.5$ millimol $\ce{H2PO4^-}$ and $1.5$ millimol $\ce{HPO4^-}$. $\endgroup$
    – Maurice
    Commented Nov 7, 2023 at 20:43
  • $\begingroup$ @Maurice Thanks. Fixed. $\endgroup$
    – Poutnik
    Commented Nov 7, 2023 at 22:26
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I got pH of 7.34, but the answer sheet says 6.97. What did I do wrong?

I'm showing here where you make the mistake as an answer to your above question (If you have already compared your answer to Poutnik's correct answer, I'm sure you'd have find where you make the mistake).

You have given following to solve the question:

The equilibrium concentrations of hydrogen phosphate and dihydrogen phosphate are $$[\ce{HPO4^2-}] = c_0(\ce{HPO4^2-}) - [\ce{H3O+}] \quad \text{and} \quad [\ce{H2PO4-}]=[\ce{H3O+}]$$ where $[\ce{H3O+}] = \frac{V(\ce{HCl})}{V(\ce{HCl)} + V(\ce{HPO4^2-)}}c_0(\ce{HCl})$.

Even though you haven't define what are the $c_0(\ce{HPO4^2-})$ and $c_0(\ce{HCl})$, I assume correctly that they are $\pu{0.20 M}$ and $\pu{0.25 M}$, respectively.

Since correct reaction is: $$\ce{HPO4^2−+H3O+ -> H2PO4^− + H2O}$$

If initial concentrations of $\ce{HPO4^2−}$ and $\ce{HCl}$ are $\alpha$ and $\beta$, and they are added in volumes $V_1$ and $V_2$, then final volume of the mixture is $(V_1 + V_2)$. Since initial $n_\ce{HPO4^2-} \gt n_\ce{HCl}$, all $\ce{HCl}$ will be converted to $\ce{H2PO4^-}$ as reaction indicated. Thus your calculation of $[\ce{H2PO4-}] = \frac{V(\ce{HCl})}{V(\ce{HCl)} + V(\ce{HPO4^2-)}} [\ce{HCl}]$ is correct before equilibrium.

However, your assumption of $[\ce{HPO4^2-}] = c_0(\ce{HPO4^2-}) - [\ce{H3O+}]$ is incorrect, simply because you have forgotten to include the fact that volume of $\ce{HPO4^2-}$ has also subjected to be diluted to $(V_1 + V_2)$ from initial volume $V_1$. Thus, your calculations gave:

$$\mathrm{pH} = \mathrm{p}K_\mathrm{a2} + \log{\frac{[\ce{HPO4^2-}]}{[\ce{H2PO4-}]}} = 7.198 + \log{\frac{0.2 - \left(\frac{0.01}{(0.01 + 0.02)}\times 0.25\right)}{\frac{0.01}{(0.01 + 0.02)}\times 0.25}} = 7.198 + 0.146 = 7.344$$ This is obviously the wrong answer. To correct it, you should have dilution factor to $\ce{HPO4^2-}$ as well:

$$\mathrm{pH} = \mathrm{p}K_\mathrm{a2} + \log{\frac{[\ce{HPO4^2-}]}{[\ce{H2PO4-}]}} = 7.198 + \log{\frac{\frac{0.02}{(0.01 + 0.02)}\times 0.20 - \left(\frac{0.01}{(0.01 + 0.02)}\times 0.25\right)}{\frac{0.01}{(0.01 + 0.02)}\times 0.25}} = 7.198 - 0.222 = 6.976$$

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