Question

The concentration of oxalic acid is ${}^{\prime }{X}^{\prime }$ mol $li{t}^{-1}.40mL$ of this solution reacts with $16mL$ of $0.05M$ acidified $KMn{O}_4$. What is the $pH$ of ${}^{\prime }{X}^{\prime }M$ oxalic acid solution?
[Assume that oxalic acid dissociates completely.]

• A. $1.3$
• B. $1.699$
• C. $1$
• D. $2$

Answer 1

Answer: (C)

$1$

$2KMn{O}_4+3H_{2}S{O}_4+5\underset{COOH}{\stackrel{COOH}{|}}\to K_{2}S{O}_4+2MnS{O}_4+8H_{2}O+10C{O}_2$

$\frac{M_1{V}_1}{n_1}\left(KMn{O}_4\right)=\frac{M_2{V}_2}{n_2}\left(\underset{COOH}{\stackrel{COOH}{|}}\right)$

$\frac{0.5\times 16}{2}=\frac{x\times 40}{5}$

$x=0.05M$

$\left[H^{+}\right]=2\times 0.05=0.1M$

$pH=-\log (H^{+}=-\log (0.1)=1$.