Answer
$pH = 8.57$
Work Step by Step
- Identify the significant electrolytes:
$Na^+$ : Insignificant
$C_7H_5{O_2}^-$: Weak base.
$K^+$: Insignificant
$Br^-$: Insignificant
- Therefore, the only reaction that will occur is the base ionization:
1. Since $C_7H_5{O_2}^-$ is the conjugate base of $HC_7H_5O_2$ , we can calculate its kb by using this equation:
$K_a * K_b = K_w = 10^{-14}$
$ 6.5\times 10^{- 5} * K_b = 10^{-14}$
$K_b = \frac{10^{-14}}{ 6.5\times 10^{- 5}}$
$K_b = 1.538\times 10^{- 10}$
2. Drawing the equilibrium (ICE) table, we get these concentrations at equilibrium:** The image is in the end of this answer.
-$[OH^-] = [HC_7H_5O_2] = x$
-$[C_7H_5{O_2}^-] = [C_7H_5{O_2}^-]_{initial} - x = 0.0887 - x$
For approximation, we consider: $[C_7H_5{O_2}^-] = 0.0887M$
3. Now, use the Kb value and equation to find the 'x' value.
$Kb = \frac{[OH^-][HC_7H_5O_2]}{ [C_7H_5{O_2}^-]}$
$Kb = 1.538 \times 10^{- 10}= \frac{x * x}{ 0.0887}$
$Kb = 1.538 \times 10^{- 10}= \frac{x^2}{ 0.0887}$
$ 1.365 \times 10^{- 11} = x^2$
$x = 3.694 \times 10^{- 6}$
Percent ionization: $\frac{ 3.694 \times 10^{- 6}}{ 0.0887} \times 100\% = 0.004165\%$
%ionization < 5% : Right approximation.
Therefore: $[OH^-] = [HC_7H_5O_2] = x = 3.694 \times 10^{- 6}M $
$[C_7H_5{O_2}^-] \approx 0.0887M$
4. Calculate the pH:
$pOH = -log[OH^-]$
$pOH = -log( 3.694 \times 10^{- 6})$
$pOH = 5.432$
$pH + pOH = 14$
$pH + 5.432 = 14$
$pH = 8.568$
