pH After titration [closed]








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> 
> $\pu{50mL}$ $\ce{SO2}$ titrated with $\pu{0.1M}$ 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 $\pu{50mL}$. Find the $\mathrm{pH}$ 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:


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


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


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.




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 $(\gt 3)$.


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