The concentration of oxalic acid is ' $x$ ' $\mathrm{mol} \mathrm{L}^{-1}$. $40 \mathrm{~mL}$ of this…
The concentration of oxalic acid is ' $x$ ' $\mathrm{mol} \mathrm{L}^{-1}$. $40 \mathrm{~mL}$ of this solution reacts with $16 \mathrm{~mL}$ of $0.05 \mathrm{M}$ acidified $\mathrm{KMnO}_4$. What is the $\mathrm{pH}$ of ' $x$ ' $\mathrm{M}$ oxalic acid solution ? (Assume that oxalic acid dissociates completely)
1.3
1.699
1
2
Solution
Oxalic acid $=x \mathrm{~mol} / \mathrm{L}$
Oxalic acid $\mathrm{KMnO}_4$
$M_1 V_1=M_2 V_2$
$40 \mathrm{~mL} \times x=16 \mathrm{~mL} \times 0.05$
$x=\frac{16 \times 0.05}{40}=\frac{1}{50}$
$x=\frac{1}{50} \mathrm{M}$
Now convert molarity into normality
$N \times$ eq. wt. $=M \times$ mol. wt. of oxalic acid
$N \times 45=\frac{1}{50} \times 90$
$N=\frac{1}{25}$
This normality represents the hydrogen ion concentration.
So, $\quad\left[\mathrm{H}^{+}ight]=\frac{1}{25}$
$\mathrm{pH}=\log \frac{1}{\left[\mathrm{H}^{+}ight]}=\log 25=1.3$