Question

Consider the three solutions of 1M concentration

I) Sodium acetate (${CH}_{3}COONa$)

II) Acetic acid + sodium acetate (${CH}_{3}COOH+{CH}_{3}COONa$)

III) Acetic acid (${CH}_{3}COOH$)

The pH of these solutions will lie in the following sequence

• A. $III<I<II$
• B. $III<II<I$
• C. $I<II<III$
• D. $II<I<III$

Answer 1

The correct option is B $III<II<I$

$1)$ $C{H}_{3}COONa$ is formed by reaction of sodium hydroxide (NaOH) and acetic acid ($C{H}_{3}COOH$). Acetic acid is a weak acid and sodium hydroxide is a strong base.

Thus, sodium acetate ($C{H}_{3}COONa$) is basic in nature and thus pH will be greater than 7

$2)$ When sodium acetate added to aq. solution of acetic acid which is acidic in nature, the concentration of $H^{+}$ decreases, $C{H}_{3}COOH$ + $C{H}_{3}COONa$ will form an acidic buffer so it's $pH<7$ but greater than $pH$ of $C{H}_{3}COOH$

$3)$ $C{H}_{3}COOH$ is a weak acid so its pH<7.

Hence, the correct option is III<II<I