What is the dissociation constant for NH_4OH if at a given temper… - Vidyadip

MCQ

What is the dissociation constant for $NH_4OH$ if at a given temperature its $0.1\,N$ solution has $pH = 11.27$ and the ionic product of water is $7.1 \times 10^{-15}$ (antilog $0.73 = 5.37$ )

A
$3\times 10^{-5}$
B
$1.86\times 10^{-6}$
✓
$1.75\times 10^{-5}$
D
$2.86\times 10^{-5}$

✓

Answer

Correct option: C.

$1.75\times 10^{-5}$

c
$\quad\left[\mathrm{H}^{+}\right]=\cdot 10^{-11.27}=10^{-12}, 10^{0.73}=5.37 \times 10^{-12}$

$\left[\mathrm{OH}^{-}\right]=\frac{\mathrm{K}_{\mathrm{w}}}{\left[\mathrm{H}^{+}\right]}=\frac{7.1 \times 10^{-15}}{5.37 \times 10^{-12}}=\frac{7.1}{5.37} \times 10^{-3}\, \mathrm{M}=\sqrt{\mathrm{K}_{\mathrm{b}} \mathrm{C}}$

$\left(\frac{7.1}{5.37}\right)^{2} \times 10^{-6}=\mathrm{K}_{\mathrm{b}} \times 0.1$

$\boxed{{K_b} = 1.747 \times {{10}^{ - 5}}}$

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