Question

$\frac{N}{10}$ acetic acid was litrated wiith $\frac{N}{10}$ sodium hydroxide. When $25\%,50\%$ and $75\%$ of nitration is over then the $pH$ of the solution will be:

$[K_a={10}^{-5}]$

• A. $5+log1/3,5,5+log3$
• B. $5+log3,4,5+log1/3$
• C. $5-log1/3,5,5-log3$
• D. $5-log1/3,4,5+log1/3$

Answer 1

The correct option is A $5+log1/3,5,5+log3$

$C{H}_{3}COOH+NaOH\to C{H}_{3}CO{O}^{-}{Na}^{+}+H_{2}O$

$n_f$ for both $C{H}_{3}COOH$ & $NaOH=1$

Therefore Normality=Molarity $0.1MC{H}_{3}COOH$ & $0.1MNaOH$

$\left[C{H}_{3}COOH\right]$ $NaOH$ $C{H}_{3}CO{O}^{-}N{a}^{+}$

$t_1=0$ $0.1$ $0.1$

$t_2$($25\%$ consumed) $0.1-0.025$ $0.1-0.025$ $0.025$

$=0.075$ $=0.075$

$t_3$( $50\%$ consumed)

$0.1-0.050$ $0.1-0.050$ $0.050$

$0.050$ $=0.050$

$t_4$( $75\%$ consumed)

$0.1-0.075$ $0.1-0.075$ $0.075$

$0.025$ $0.025$

$P_H=-\log K_a+\log \frac{Salt}{Acid}$ $-\log K_a=-\log {10}^{-5}$

$=5\log 10$

$=5$

For $25\%$ completion

$P_H=5+\log \frac{0.025}{0.075}$

$=5+\log \frac{1}{3}$

For $50\%$ completion

$P_H=5+\log \frac{0.050}{0.050}$

$=5+\log 1$

$=5$

For $75\%$ completion

$P_H=5+\log \frac{0.075}{0.025}$

$=5+\log 3$