Question

The pH of a solution of $N{H}_3$ is 11.612. If its concentration is 0.95 M then what is the value of its dissociation constant ?

• A. Anti log [28 + log (0.95) - 23.242]
• B. Anti log [23.242 - log (0.95) - 28]
• C. Anti log [11.612 - log (0.95) - 14]
• D. Anti log [14 + log (0.95) - 11.612]

Answer 1

The correct option is B

Anti log [23.242 - log (0.95) - 28]

Since $pH$ = $14-pOH$ and $pOH$

= $\frac{1}{2}p{K}_b$ - $\frac{1}{2}logC$

or $pH$ = $14-\frac{1}{2}p{K}_b+\frac{1}{2}logC$

or $p{K}_b$ = $2(14+\frac{1}{2}logC-pH)$

or $K_b$ = $antilog[23.242-log(0.95)-28]$