Answer
The pH change is equal to $3.78$.
Work Step by Step
1. Find the concentration after the dilution:
$C_1 * V_1 = C_2 * V_2$
$1*1= C_2 *100$
$1 = C_2 *100$
$C_2 =0.01$
2. Drawing the ICE table we get these concentrations at the equilibrium:
$CH_3COOH(aq) + H_2O(l) \lt -- \gt CH_3COO^-(aq) + H_3O^+(aq)$
Remember: Reactants at equilibrium = Initial Concentration - x
And Products = Initial Concentration + x
$[CH_3COOH] = 0.01 M - x$
$[CH_3COO^-] = 0.01M + x$
$[H_3O^+] = 0 + x$
3. Calculate 'x' using the $K_a$ expression.
$ 1.8\times 10^{- 5} = \frac{[CH_3COO^-][H_3O^+]}{[CH_3COOH]}$
$ 1.8\times 10^{- 5} = \frac{( 0.01 + x )* x}{ 0.01 - x}$
Considering 'x' has a very small value.
$ 1.8\times 10^{- 5} = \frac{ 0.01 * x}{ 0.01}$
$ 1.8\times 10^{- 5} = 1x$
$\frac{ 1.8\times 10^{- 5}}{ 1} = x$
$x = 1.8\times 10^{- 5}$
Percent dissociation: $\frac{ 1.8\times 10^{- 5}}{ 0.01} \times 100\% = 0.18\%$
x = $[H_3O^+]$
$[CH_3COOH] = 0.01 M - x = 0.01 M - 1.8 \times 10^{-5}M \approx 0.01M$
$[CH_3COO^-] = 0.01M + x = 0.01 M + 1.8 \times 10^{-5}M \approx 0.01M$
$[H_3O^+] = 0 + x = 1.8 \times 10^{-5}M$
4. Calculate the pH Value
$pH = -log[H_3O^+]$
$pH = -log( 1.8 \times 10^{- 5})$
$pH = 4.74$
5. Since we are adding a strong base, this reaction will occur:
$CH_3COOH(aq) + OH^-(aq) -- \gt CH_3COO^-(aq) + H_2O(l)$
And these are the concentrations after this reaction:
Remember: Reactants at equilibrium = Initial Concentration - y
And Products = Initial Concentration + y
Since $NaOH$ is a strong base, y = $[NaOH] = 0.01M$
$[CH_3COOH] = 0.01 M - 0.01 = 0M$
$[CH_3COO^-] = 0.01M + 0.01 = 0.02M$
- The only compound that reacts in the solution is $CH_3COO^-$ with 0.02M
So, calculate the pH of a solution with it.
- 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}$
$K_b = \frac{10^{-14}}{ 1.8\times 10^{- 5}}$
$K_b = 5.6\times 10^{- 10}$
6. We have these concentrations at equilibrium:
-$[OH^-] = [CH_3COOH] = x$
-$[CH_3COO^-] = [CH_3COO^-]_{initial} - x = 0.02 - x$
For approximation, we consider: $[CH_3COO^-] = 0.02M$
7. Now, use the Kb value and equation to find the 'x' value.
$Kb = \frac{[OH^-][CH_3COOH]}{ [CH_3COO^-]}$
$Kb = 5.6 \times 10^{- 10}= \frac{x * x}{ 0.02}$
$Kb = 5.6 \times 10^{- 10}= \frac{x^2}{ 0.02}$
$ 1.1 \times 10^{- 11} = x^2$
$x = 3.3 \times 10^{- 6}$
Percent ionization: $\frac{ 3.3 \times 10^{- 6}}{ 0.02} \times 100\% = 0.017\%$
%ionization < 5% : Right approximation.
Therefore: $[OH^-] = [CH_3COOH] = x = 3.3 \times 10^{- 6}M $
$[CH_3COO^-] \approx 0.02M$
8. Calculate the pOH:
$pOH = -log[OH^-]$
$pOH = -log( 3.3 \times 10^{- 6})$
$pOH = 5.48$
9. Find the pH:
$pH + pOH = 14$
$pH + 5.48 = 14$
$pH = 8.52$
10. Calculate the pH change:
$\Delta pH = 8.52 - 4.74 = 3.78$
