Question

In which of the following mixed aqueous solutions $pH=pKa$ at equilibrium?

$\left(1\right)100ml$ of $0.1MC{H}_{3}COOH+100ml$ of $0.1MC{H}_{3}COONa$
$\left(2\right)100ml$ of $0.1MC{H}_{3}COOH+50ml$ of $0.1MNaOH$
$\left(3\right)100ml$ of $0.1MC{H}_{3}COOH+100ml$ of $0.1MNaOH$
$\left(4\right)100ml$ of $0.1MC{H}_{3}COOH+100ml$ of $0.1MN{H}_3$

• A. $\left(1\right)$ is correct
• B. $\left(2\right)$ is correct
• C. $\left(3\right)$ is correct
• D. Both $\left(1\right)$ and $\left(2\right)$ are correct

Answer 1

Answer: (D)

Both $\left(1\right)$ and $\left(2\right)$ are correct

Both (1) and (2) are correct.
$pH=p{K}_a$ at equilibrium for the (1) and (2) aqueous solutions.
(1) A solution containing a weak acid and its salt with strong base is an acidic buffer solution.
$pH=p{K}_a+\log \frac{\left[C{H}_{3}COONa\right]}{\left[C{H}_{3}COOH\right]}$
$pH=p{K}_a+\log \left(\frac{0.1\times 100}{0.1\times 100}\right)=p{K}_a$ (correct).
(2) A solution containing a weak acid and a strong base is an acidic buffer solution.

$\underset{\underset{5mmol}{10mmol}}{C{H}_{3}COOH}+\underset{\underset{0}{5mmol}}{NaOH}\to \underset{\underset{5mmol}{0}}{C{H}_{3}CO{O}^{-}}N{a}^{+}+\underset{\underset{5mmol(\text{acidic buffer formed)}}{0}}{H_{2}O}$
$pH=p{K}_a+\log \frac{\left[C{H}_{3}COONa\right]}{\left[C{H}_{3}COOH\right]}$
$pH=p{K}_a+\log \left(\frac{5}{5}\right)=p{K}_a$ (Correct).
(3) A salt of weak acid (acetic acid) and strong base (NaOH) is formed. It undergoes hydrolysis.
$\underset{\underset{0}{10mmol}}{C{H}_{3}COOH}+\underset{\underset{0}{10mmol}}{NaOH}\to \underset{\underset{10mmol}{0}}{C{H}_{3}CO{O}^{-}}N{a}^{+}+\underset{\underset{10mmol}{0}}{H_{2}O}$
$pH=\frac{1}{2}(p{K}_w+p{K}_a+\log c)=\frac{1}{2}[14+p{K}_a+\log \left(\frac{10}{200}\right)]$ (incorrect) (anionic hydrolsis).

(4) Weak acid and weak base give buffer solution.

$C{H}_{3}COOH+N{H}_{4}OH\to C{H}_{3}CO{O}^{-}+N{H}_4^{+}$

10 mmol 10mmol 0 0

0 0 10mmol 10mmol

In this case, pH won't be equal to pKa because in the weak acid and weak base titration pKb will also come into the picture from the following formula:
$pH=\frac{1}{2}p{K}_w+\frac{1}{2}p{K}_a-\frac{1}{2}p{K}_b$

Hence the correct option is