Question

An aqueous solution contains 10% ammonia by mass and has a density 0.99 g cm-3. If Ka for NH₄⁺ is 5.0×10-10 M, the value of hydroxyl ion concentration will be

• A. 9.27×10^-13
• B. 9.27×10^-11
• C. 9.27×10^-10
• D. 9.27×10^-6

Answer 1

The correct option is A

9.27×10^-13

Mass of $1L$ solution $=1000\times 0.99g=990g$
${NH}_3$ present in $990g$ solution $=10\%$ of $990=99g$

$1L$ solution contains $=\frac{99}{17}=5.8mol{NH}_3$
${NH}_3+H_{2}O\rightleftharpoons {NH}_{4}OH={NH}_4^{+}+{OH}^{-}$

$\begin{array}{cccc}\text{At}t=0& 5.8& 0& 0\\ \text{At equilibrium}& (5.8-X)& x& x\end{array}$

$\therefore K_b=\frac{\left|N{H}_4^{+}\right|O{H}^{-}\mid }{\left[N{H}_3\right]}=\frac{xx}{(5.8-x)}\approx \frac{x^2}{5.8}\dots \dots \dots .\left(t\right)$

Again $K_b=\frac{K_w}{K_a}=\frac{{10}^{-14}}{5.0\times {10}^{-10}}=2\times {10}^{-5}$
From Eqs. $\left(i\right)$ and $\left(ii\right)$ we get
$\frac{x^2}{5.8}=2\times {10}^{-5}$

$x^2=5.8\times 2\times {10}^{-5}$

i.e., $\left[O{H}^{-}\right]=x=1.078\times {10}^{-2}M$

$\because \left[H^{\ast }\right]\left[O{H}^{-}\right]={10}^{-14}$
$\therefore \left[H^{+}\right]=\frac{{10}^{-14}}{1.078\times {10}^{-12}}=9.27\times {10}^{-13}M$