Question

$\frac{N}{10}$ acetic acid was titrated with $\frac{N}{10}$ $NaOH$. when 25%, 50% and 75% of titration is over then the $pH$ of the solution will be (respectively):

$[K_a={10}^5]$

• A. $5+log1/3,5,5+log3$
• B. $5+log3,4,5,+1/3$
• C. $5-log1/3,5,5-log3$
• D. $5-log1/3,4,5+log1/3$

Answer 1

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

$n_f$ for both ${CH}_{3}COOH$ and $NaOH=1$

$\therefore$ Normality $=$ Molarilty

$\frac{N}{10}=\frac{1}{10}=0.1H$ acetic acid

& $0.1H$ $NaOH$ also

${CH}_{3}COOH+NaOH\rightleftharpoons {CH}_3{COO}^{(-)}{Na}^{(+)}+H_{2}O$

initial $0.1$ $0.1$ $0$ $0$

equilibrium $0.075$ $0.075$ $0.025$ $0.025$

at $t_2$

Since, $25$% of $0.1$ is consumed

at $t_3$ $0.1-0.050$ $0.1-0.050$ $0.050$

($50$% of $0.1$ $=0.050$ $=0.50$

is consumed)

at $t_4$ $0.1-0.075$ $0.1-0.075$ $0.075$

($75$% of $0.1$ $=0.025$ $=0.025$

is consumed)

$pH=-log\left(K_a\right)+log\left(\frac{Salt}{Acid}\right)$

For $25$% completion

$pH=-log{K}_a+log\left(\frac{0.025}{0.075}\right)$

$=5+log\left(1/3\right)$

For $50$% completion

$pH=-log{K}_a+log\left(\frac{0.050}{0.050}\right)$

$=5+0=5$

For $75$% completion

$pH=-log{K}_a+log\left(\frac{0.075}{0.025}\right)$

$=5+log3$

$\Rightarrow$ Option A) $5+log\left(1/3\right),5,5+log3$