50mL $\ce{SO2}$ titrated with 0.1M of $\ce{KBrO4}$ with reaction: $$\ce{KBrO4 + 4SO2 + H2O -> 4H2SO4 + KBr}$$ The equivalent point is reached when the volume of $\ce{KBrO4}$ is used as much 50mL. Find the pH of the solution after titration.

(A) $4$

(B) $4-\log_2$

(C) $4-2\log_2$

(D) $5$

(E) $6$

Here's my attempt to solve the problem:

$$\begin{aligned}n_{\ce{KBrO4}}&=M\cdot V\\&=5mmol \\\\ n_{\ce{H2SO4}}&=4n_{\ce{KBrO4}}\\ &=20mmol \end{aligned}$$

$$\ce{H2SO4-> 2H+ + SO4^{2-}}$$ $$\begin{aligned} n_{\ce{H+}}&=2n_{\ce{H2SO4}}\\ &=40mmol \\ [H^+]&=\frac{40mmol}{100mL}\\ &=0.4M \\ \\ pH&=1-2\log2 \end{aligned}$$

It's not on the option. I thought I might forget to add the water volume, but there's no further information about the water volume.

  • $\begingroup$ If it is correct that the solution contains about 0.2 M sulfuric acid, all the pH values in the multiple choice answers are too high. Maybe the question asked about 0.1 mM KBrO4? $\endgroup$ – Karsten Theis Apr 29 at 17:28
  • 1
    $\begingroup$ 50 mL of SO2 ??? It was a gas in my student days, with the boiling point -10 Deg C. $\endgroup$ – Poutnik Apr 29 at 18:43

Based on the quality of the question, only I can try to speculate. As Poutnik pointed out that $\ce{SO2}$ is a gas, but dissolve in water very much, so the question must be talking about $\pu{50 mL}$ of hydrated $\ce{SO2}$. That's probably where water in the equation is coming from. Yet, the equation itself is not balanced. The balanced equation is as follows:

$$\ce{KBrO4 + 4SO2 + 4 H2O -> KBr + 4 H2SO4}$$

As calculated by OP, final $\ce{[H+]}$ is $\pu{0.4 mol/L}$ (or $\pu{0.4 M}$)

Thus, $\mathrm{pH} = -\log \ce{[H+]}= 0.4 = 1-2\log 2$ (OP's final answer is correct; my apology to OP for earlier mishap!).

This means Karsten Theis is correct: All the $\mathrm{pH}$ values in the multiple choice answers are too high ($\gt 3$).

  • 2
    $\begingroup$ It seems to me the author of that task should be downvoted. :-) $\endgroup$ – Poutnik Apr 30 at 5:00

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