Answer
$[OH^-] = 9.177 \times 10^{- 5}M $
$pH = 9.96$
Work Step by Step
- Identify the significant electrolytes in $NaHS$:
- $Na^+$: Insignificant acid.
- $HS^-$: Weak acid.
Therefore, we just need to calculate the $pH$ for a $0.080M$ $HS^-$ solution:
1. Since $HS^-$ is the conjugate base of $H_2S$ , we can calculate its kb by using this equation:
$K_a * K_b = K_w = 10^{-14}$
$ 9.5\times 10^{- 8} * K_b = 10^{-14}$
$K_b = \frac{10^{-14}}{ 9.5\times 10^{- 8}}$
$K_b = 1.053\times 10^{- 7}$
2. Drawing the equilibrium (ICE) table, we get these concentrations at equilibrium:** The image is in the end of this answer.
-$[OH^-] = [H_2S] = x$
-$[HS^-] = [HS^-]_{initial} - x = 0.08 - x$
For approximation, we consider: $[HS^-] = 0.08M$
3. Now, use the Kb value and equation to find the 'x' value.
$Kb = \frac{[OH^-][H_2S]}{ [HS^-]}$
$Kb = 1.053 \times 10^{- 7}= \frac{x * x}{ 0.08}$
$Kb = 1.053 \times 10^{- 7}= \frac{x^2}{ 0.08}$
$ 8.421 \times 10^{- 9} = x^2$
$x = 9.177 \times 10^{- 5}$
Percent ionization: $\frac{ 9.177 \times 10^{- 5}}{ 0.08} \times 100\% = 0.1147\%$
%ionization < 5% : Right approximation.
Therefore: $[OH^-] = [H_2S] = x = 9.177 \times 10^{- 5}M $
$[HS^-] \approx 0.08M$
4. Calculate the pH.
$pOH = -log[OH^-]$
$pOH = -log( 9.177 \times 10^{- 5})$
$pOH = 4.037$
$pH + pOH = 14$
$pH + 4.037 = 14$
$pH = 9.96$
