Question

What will be the new $pH$ if $0.01mol$ of $NaOH$ is added to a $1L$ buffer solution that is $0.1M$ in acetic acid $\left(C{H}_{3}COOH\right)$ and $0.1M$ in sodium acetate $\left(C{H}_{3}COONa\right)$ ?
Dissociation constant $K_a$ of acetic acid at ${25}^{\circ }C$ is $1.8\times {10}^{-5}$

• A. 5.52
• B. 4.92
• C. 4.1
• D. 6.13

Answer 1

The correct option is B 4.92

$p{K}_a=-\log \left(K_a\right)p{K}_a=-\log (1.8\times {10}^{-5})p{K}_a=(5-0.26)=4.74$
Finding initial $pH$ of the buffer
$pH$ for acidic buffer can be defined by :

$pH=p{K}_a+\log \left(\frac{\left[C{H}_{3}CO{O}^{-}\right]}{\left[C{H}_{3}COOH\right]}\right)pH=4.74+\log \left(\frac{0.1}{0.1}\right)pH=4.74+\log \left(1\right)pH=4.74$

Since, Volume = $1L$
Concentration = moles
$\therefore \left[NaOH\right]=0.02M$

On addition of small amount of base ($NaOH$) externally there have a reaction like,
$C{H}_{3}COOH+NaOH\rightleftharpoons C{H}_{3}COONa+H_{2}O$.
So, the concentration of conjugate base increases whereas concentration of acid decreases by same amount.

So,
$\left[C{H}_{3}CO{O}^{-}\right]=0.1+0.02=0.12M\left[C{H}_{3}COOH\right]=0.1-0.02=0.08M$

Finding the changed $pH$ :

$p{H}_{new}=p{K}_a+\log \left(\frac{\left[C{H}_{3}CO{O}^{-}\right]}{\left[C{H}_{3}COOH\right]}\right)p{H}_{new}=4.74+\log \left(\frac{0.12}{0.08}\right)p{H}_{new}=4.74+\log \left(\frac{3}{2}\right)p{H}_{new}=4.74+\log \left(3\right)-\log \left(2\right)p{H}_{new}=4.74+0.48-0.30=4.92$