Answer
The pH of this solution is going to be equal to $8.81$.
Work Step by Step
- Find the numbers of moles:
$C(CH_3COOH) * V(CH_3COOH) = 0.15* 0.035 = 5.25 \times 10^{-3}$ moles
$C(NaOH) * V(NaOH) = 0.15* 0.035 = 5.25 \times 10^{-3}$ moles
When the number of moles is equal, the reactants are totally consumed:
$CH_3COOH(aq) + NaOH(aq) -- \gt NaCH_3COO(aq) + H_2O(l)$
- Total volume: 0.035 + 0.035 = 0.07L
- So, those are the final concentrations:
$[CH_3COOH] = 0.00525 - 0.00525 = 0$ mol.
$[NaOH] = 0.00525 - 0.00525 = 0$ mol
$[NaCH_3COO] = 0 + 0.00525 = 0.00525$ moles.
Concentration: $\frac{ 0.00525}{ 0.07} = 0.075M$
- Therefore, we have a weak base salt solution:
- Since $CH_3COO^-$ is the conjugate base of $CH_3COOH$ , we can calculate its kb by using this equation:
$K_a * K_b = K_w = 10^{-14}$
$ 1.8\times 10^{- 5} * K_b = 10^{-14}$
$K_b = \frac{10^{-14}}{ 1.8\times 10^{- 5}}$
$K_b = 5.556\times 10^{- 10}$
- We have these concentrations at equilibrium:
-$[OH^-] = [CH_3COOH] = x$
-$[CH_3COO^-] = [CH_3COO^-]_{initial} - x = 0.075 - x$
For approximation, we consider: $[CH_3COO^-] = 0.075M$
- Now, use the Kb value and equation to find the 'x' value.
$Kb = \frac{[OH^-][CH_3COOH]}{ [CH_3COO^-]}$
$Kb = 5.556 \times 10^{- 10}= \frac{x * x}{ 0.075}$
$Kb = 5.556 \times 10^{- 10}= \frac{x^2}{ 0.075}$
$ 4.167 \times 10^{- 11} = x^2$
$x = 6.455 \times 10^{- 6}$
Percent ionization: $\frac{ 6.455 \times 10^{- 6}}{ 0.075} \times 100\% = 0.008607\%$
%ionization < 5% : Right approximation.
Therefore: $[OH^-] = [CH_3COOH] = x = 6.455 \times 10^{- 6}M $
$[CH_3COO^-] \approx 0.075M$
$pOH = -log[OH^-]$
$pOH = -log( 6.455 \times 10^{- 6})$
$pOH = 5.19$
$pH + pOH = 14$
$pH + 5.19 = 14$
$pH = 8.81$
