Answer
The pH change is equal to $0.08$.
** Considering only one significant digit in the pH:
The pH change is equal to $0.1$
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 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
$[CH_3COOH] = 0.1 M - x$
$[CH_3COO^-] = 0.1M + x$
$[H_3O^+] = 0 + x$
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{( 0.1 + x )* x}{ 0.1 - x}$
Considering 'x' has a very small value.
$ 1.8\times 10^{- 5} = \frac{ 0.1 * x}{ 0.1}$
$ 1.8\times 10^{- 5} = 1x$
$\frac{ 1.8\times 10^{- 5}}{ 1} = x$
$x = 1.8\times 10^{- 5}$
Percent dissociation: $\frac{ 1.8\times 10^{- 5}}{ 0.1} \times 100\% = 0.018\%$
x = $[H_3O^+]$
$[CH_3COOH] = 0.1 M - x = 0.1 M - 1.8 \times 10^{-5}M \approx 0.1M$
$[CH_3COO^-] = 0.1M + x = 0.1 M + 1.8 \times 10^{-5}M \approx 0.1M$
$[H_3O^+] = 0 + x = 1.8 \times 10^{-5}M$
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 = $[NaOH] = 0.01M$
$[CH_3COOH] = 0.1 M - 0.01 = 0.09M$
$[CH_3COO^-] = 0.1M + 0.01 = 0.11M$
6. Now, calculate the hydronium ion concentration after the addition of the $NaOH$:
$[H_3O^+] = Ka * (\frac{[CH_3COOH]}{[CH_3COO^-]})$
$[H_3O^+] = 1.8 \times 10^{-5} * \frac{0.09}{0.11}$
$[H_3O^+] = 1.8 \times 10^{-5} * 0.82$
$[H_3O^+] = 1.5 \times 10^{-5}$
7. Calculate the pH Value
$pH = -log[H_3O^+]$
$pH = -log( 1.5 \times 10^{- 5})$
$pH = 4.82$
8. Find the pH change:
$\Delta pH = 4.82 - 4.74 = 0.08$
** Considering one significant digit in pH:
$\Delta pH = 4.8 - 4.7 = 0.1$
