Answer
$pH = 4.99$
Work Step by Step
1. Find the concentration after the dilution:
$C_1 * V_1 = C_2 * V_2$
$0.1*1= C_2 *10$
$0.1 = C_2 *10$
$C_2 =0.01$
2. Drawing the ICE table we get these concentrations at the equilibrium:
$C_3H_5O_2H(aq) + H_2O(l) \lt -- \gt C_3H_5O{_2}^-(aq) + H_3O^+(aq)$
Remember: Reactants at equilibrium = Initial Concentration - x
And Products = Initial Concentration + x
$[C_3H_5O_2H] = 0.2 M - x$
$[C_3H_5O{_2}^-] = 0.3M + x$
$[H_3O^+] = 0 + x$
3. Calculate 'x' using the $K_a$ expression.
$ 1.4\times 10^{- 5} = \frac{[C_3H_5O{_2}^-][H_3O^+]}{[C_3H_5O_2H]}$
$ 1.4\times 10^{- 5} = \frac{( 0.3 + x )* x}{ 0.2 - x}$
Considering 'x' has a very small value.
$ 1.4\times 10^{- 5} = \frac{ 0.3 * x}{ 0.2}$
$ 1.4\times 10^{- 5} = 1.5x$
$\frac{ 1.4\times 10^{- 5}}{ 1.5} = x$
$x = 9.333\times 10^{- 6}$
Percent dissociation: $\frac{ 9.333\times 10^{- 6}}{ 0.2} \times 100\% = 4.667\times 10^{- 3}\%$
x = $[H_3O^+]$
4. Calculate the pH Value
$pH = -log[H_3O^+]$
$pH = -log( 9.333 \times 10^{- 6})$
$pH = 5.03$
5. Since we are adding a strong acid, this reaction will occur:
$C_3H_5O{_2}^-(aq) + H_3O^+(aq) -- \gt C_3H_5O_2H(aq) + H_2O(l)$
And these are the concentrations after this reaction:
Remember: Reactants at equilibrium = Initial Concentration - y
And Products = Initial Concentration + y
Since $HCl$ is a strong acid, y = $[HCl] = 0.01M$
$[C_3H_5O_2H] = 0.2 M + 0.01 = 0.21M$
$[C_3H_5O{_2}^-] = 0.3M - 0.01 = 0.28M$
6. Now, calculate the hydronium ion concentration after the addition of the $HCl$:
$[H_3O^+] = Ka * (\frac{[C_3H_5O_2H]}{[C_3H_5O{_2}^-]})$
$[H_3O^+] = 1.4 \times 10^{-5} * \frac{0.21}{0.28}$
$[H_3O^+] = 1.4 \times 10^{-5} * 0.7241$
$[H_3O^+] = 1.014 \times 10^{-5}$
7. Calculate the pH Value
$pH = -log[H_3O^+]$
$pH = -log( 1.014 \times 10^{- 5})$
$pH = 4.99$
