Answer
The pH of this solution is equal to $10.98$
Work Step by Step
- This solution is only made of $0.050M$ $NH_3$, so we can use its $K_b$ to calculate the pH:
- We have these concentrations at equilibrium:
-$[OH^-] = [N{H_4}^+] = x$
-$[NH_3] = [NH_3]_{initial} - x = 0.05 - x$
For approximation, we consider: $[NH_3] = 0.05M$
- Now, use the Kb value and equation to find the 'x' value.
$Kb = \frac{[OH^-][N{H_4}^+]}{ [NH_3]}$
$Kb = 1.8 \times 10^{- 5}= \frac{x * x}{ 0.05}$
$Kb = 1.8 \times 10^{- 5}= \frac{x^2}{ 0.05}$
$ 9 \times 10^{- 7} = x^2$
$x = 9.487 \times 10^{- 4}$
Percent ionization: $\frac{ 9.487 \times 10^{- 4}}{ 0.05} \times 100\% = 1.897\%$
%ionization < 5% : Right approximation.
Therefore: $[OH^-] = [N{H_4}^+] = x = 9.487 \times 10^{- 4}M $
$[NH_3] \approx 0.05M$
$pOH = -log[OH^-]$
$pOH = -log( 9.487 \times 10^{- 4})$
$pOH = 3.023$
$pH + pOH = 14$
$pH + 3.023 = 14$
$pH = 10.977$
