Answer
$pH = 8.31$
Work Step by Step
- Find the numbers of moles:
$C(C_6H_5COOH) * V(C_6H_5COOH) = 0.1* 0.03 = 3 \times 10^{-3}$ moles
$C(NaOH) * V(NaOH) = 0.1* 0.03 = 3 \times 10^{-3}$ moles
When the number of moles is equal, the reactants are totally consumed:
$C_6H_5COOH(aq) + NaOH(aq) -- \gt NaC_6H_5COO(aq) + H_2O(l)$
- Total volume: 0.03 + 0.03 = 0.06L
- So, these are the final concentrations:
$[C_6H_5COOH] = 0.003 - 0.003 = 0$ mol.
$[NaOH] = 0.003 - 0.003 = 0$ mol
$[NaC_6H_5COO] = 0 + 0.003 = 0.003$ moles.
Concentration: $\frac{ 0.003}{ 0.06} = 0.05M$
- Therefore, we have a weak base salt solution:
- Since $C_6H_5COO^-$ is the conjugate base of $C_6H_5COOH$, we can calculate its kb by using this equation:
$K_a * K_b = K_w = 10^{-14}$
$ 1.2\times 10^{- 4} * K_b = 10^{-14}$
$K_b = \frac{10^{-14}}{ 1.2\times 10^{- 4}}$
$K_b = 8.333\times 10^{- 11}$
- We have these concentrations at equilibrium:
-$[OH^-] = [C_6H_5COOH] = x$
-$[C_6H_5COO^-] = [C_6H_5COO^-]_{initial} - x = 0.05 - x$
For approximation, we consider: $[C_6H_5COO^-] = 0.05M$
- Now, use the Kb value and equation to find the 'x' value.
$Kb = \frac{[OH^-][C_6H_5COOH]}{ [C_6H_5COO^-]}$
$Kb = 8.333 \times 10^{- 11}= \frac{x * x}{ 0.05}$
$Kb = 8.333 \times 10^{- 11}= \frac{x^2}{ 0.05}$
$ 4.167 \times 10^{- 12} = x^2$
$x = 2.041 \times 10^{- 6}$
Percent ionization: $\frac{ 2.041 \times 10^{- 6}}{ 0.05} \times 100\% = 0.004082\%$
%ionization < 5% : Right approximation.
Therefore: $[OH^-] = [C_6H_5COOH] = x = 2.041 \times 10^{- 6}M $
$[C_6H_5COO^-] \approx 0.05M$
$pOH = -log[OH^-]$
$pOH = -log( 2.041 \times 10^{- 6})$
$pOH = 5.69$
$pH + pOH = 14$
$pH + 5.69 = 14$
$pH = 8.31$
