# pH After Titration

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.

• 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? – Karsten Theis Apr 29 at 17:28
• 50 mL of SO2 ??? It was a gas in my student days, with the boiling point -10 Deg C. – 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 ($$\gt 3$$).