Question

What is the difference in pH for 1/3 and 2/3 stages of neutralization of 0.1 M $C{H}_{3}COOH$ with 0.1 M $NaOH$ ?

• A. log 9
• B. 2 log (1/4)
• C. 2 log (2/3)
• D. log 4

Answer 1

The correct option is D log 4

$C{H}_{3}COOH\rightleftharpoons C{H}_{3}CO{O}^{-}+H^{+}$

$K_a=1.8\times {10}^{-5}$

$pH=p{K}_a+\log \frac{\left[C{H}_{3}CO{O}^{-}\right]}{\left[C{H}_{3}COOH\right]}$

In the $1/3$ stage of neutralization,

$1/3$ of $C{H}_{3}COOH$ is changed to

$C{H}_{3}CO{O}^{-}$

Hence, $\left[C{H}_{3}CO{O}^{-}\right]/\left[C{H}_{3}COOH\right]=\left(1/3\right)/(1-1/3)=1/2$

$\therefore p{H}_1=-\log (1.8\times {10}^{-5})+\log 1/2\to 1$

In the $2/3$ stage of neutralization,

$\left[C{H}_{3}CO{O}^{-}\right]/\left[C{H}_{3}COOH\right]=\left(2/3\right)/(1-2/3)=2$

$\therefore p{H}_2=-\log (1.8\times {10}^{-5})+\log 2\to 2$

$p{H}_2-p{H}_1=\log 2-\log 1/2=2\log 2-\log 1=\log 4-\log 1$

$=\log 4$