Answer
$[OH^-] = 3.3 \times 10^{- 6}M $
$pH = 8.52$
Work Step by Step
- Identify the significant electrolytes in the solution:
- $0.10M$ $Na^+$: Insignificant
- $0.10M$ $N{O_2}^-$: Weak base
- $0.20M$ $Ca^{2+}$: Insignificant
- $0.40M$** $N{O_2}^-$: Weak base
** 0.20M * 2 (Number of $N{O_2}$ in $Ca(NO_3)_2$
0.10M + 0.40M =$ 0.50M$ $N{O_2}^-$
Calculate the pH of a $ 0.50M$ $N{O_2}^-$ solution.
1. Since $N{O_2}^-$ is the conjugate base of $HNO_2$ , we can calculate its kb by using this equation:
$K_a * K_b = K_w = 10^{-14}$
$ 4.5\times 10^{- 4} * K_b = 10^{-14}$
$K_b = \frac{10^{-14}}{ 4.5\times 10^{- 4}}$
$K_b = 2.222\times 10^{- 11}$
2. Drawing the equilibrium (ICE) table, we get these concentrations at equilibrium:** The image is in the end of this answer.
-$[OH^-] = [HNO_2] = x$
-$[N{O_2}^-] = [N{O_2}^-]_{initial} - x = 0.5 - x$
For approximation, we consider: $[N{O_2}^-] = 0.5M$
3. Now, use the Kb value and equation to find the 'x' value.
$Kb = \frac{[OH^-][HNO_2]}{ [N{O_2}^-]}$
$Kb = 2.222 \times 10^{- 11}= \frac{x * x}{ 0.5}$
$Kb = 2.222 \times 10^{- 11}= \frac{x^2}{ 0.5}$
$ 1.111 \times 10^{- 11} = x^2$
$x = 3.333 \times 10^{- 6}$
Percent ionization: $\frac{ 3.333 \times 10^{- 6}}{ 0.5} \times 100\% = 0.0006667\%$
%ionization < 5% : Right approximation.
Therefore: $[OH^-] = [HNO_2] = x = 3.333 \times 10^{- 6}M $
$[N{O_2}^-] \approx 0.5M$
4. Calculate the pH:
$pOH = -log[OH^-]$
$pOH = -log( 3.333 \times 10^{- 6})$
$pOH = 5.477$
$pH + pOH = 14$
$pH + 5.477 = 14$
$pH = 8.523$
