Answer
(a) $pH = 4.95$
(b) $pH = 5.05$
Work Step by Step
1. Calculate the molar mass $(NaOH)$:
22.99* 1 + 16* 1 + 1.01* 1 = 40g/mol
2. Calculate the number of moles $(NaOH)$
$n(moles) = \frac{mass(g)}{mm(g/mol)}$
$n(moles) = \frac{ 0.082}{ 40}$
$n(moles) = 2.1\times 10^{- 3}$
3. Find the concentration in mol/L $(NaOH)$:
$C(mol/L) = \frac{n(moles)}{volume(L)}$
$ C(mol/L) = \frac{ 2.1\times 10^{- 3}}{ 0.1} $
$C(mol/L) = 0.021$
4. Calculate the molar mass $(NaCH_3CO_2)$:
22.99* 1 + 12.01* 1 + 1.01* 3 + 12.01* 1 + 16* 2 = 82.04g/mol
5. Calculate the number of moles $(NaCH_3CO_2)$
$n(moles) = \frac{mass(g)}{mm(g/mol)}$
$n(moles) = \frac{ 4.95}{ 82.04}$
$n(moles) = 0.06$
6. Find the concentration in mol/L $(NaCH_3CO_2)$:
$C(mol/L) = \frac{n(moles)}{volume(L)}$
$ C(mol/L) = \frac{ 0.06}{ 0.25} $
$C(mol/L) = 0.24$
7. Drawing the ICE table we get these concentrations at the equilibrium:
$CH_3CO_2H(aq) + H_2O(l) \lt -- \gt CH_3C{O_2}^-(aq) + H_3O^+(aq)$
Remember: Reactants at equilibrium = Initial Concentration - x
And Products = Initial Concentration + x
$[CH_3CO_2H] = 0.15 M - x$
$[CH_3C{O_2}^-] = 0.24M + x$
$[H_3O^+] = 0 + x$
8. Calculate 'x' using the $K_a$ expression.
$ 1.8\times 10^{- 5} = \frac{[CH_3C{O_2}^-][H_3O^+]}{[CH_3CO_2H]}$
$ 1.8\times 10^{- 5} = \frac{( 0.24 + x )* x}{ 0.15 - x}$
Considering 'x' has a very small value.
$ 1.8\times 10^{- 5} = \frac{ 0.24 * x}{ 0.15}$
$ 1.8\times 10^{- 5} = 1.6x$
$\frac{ 1.8\times 10^{- 5}}{ 1.6} = x$
$x = 1.13\times 10^{- 5}$
Percent dissociation: $\frac{ 1.13\times 10^{- 5}}{ 0.15} \times 100\% = 7.5\times 10^{- 3}\%$
x = $[H_3O^+]$
$[CH_3CO_2H] = 0.15 M - x = 0.15 M - 1.13 \times 10^{-5}M \approx 0.15M$
$[CH_3C{O_2}^-] = 0.24M + x = 0.15 M + 1.13 \times 10^{-5}M \approx 0.24M$
$[H_3O^+] = 0 + x = 1.13 \times 10^{-5}M$
9. Calculate the pH Value
$pH = -log[H_3O^+]$
$pH = -log( 1.13 \times 10^{- 5})$
$pH = 4.95$
10. Since we are adding a strong base, this reaction will occur:
$CH_3CO_2H(aq) + OH^-(aq) -- \gt CH_3C{O_2}^-(aq) + H_2O(l)$
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 = $[NaOH] = 0.021M$
$[CH_3CO_2H] = 0.15 M - 0.021 = 0.13M$
$[CH_3C{O_2}^-] = 0.24M + 0.021 = 0.26M$
11. Now, calculate the hydronium ion concentration after the addition of the $NaOH$:
$[H_3O^+] = Ka * (\frac{[CH_3CO_2H]}{[CH_3C{O_2}^-]})$
$[H_3O^+] = 1.8 \times 10^{-5} * \frac{0.13}{0.26}$
$[H_3O^+] = 1.8 \times 10^{-5} * 0.49$
$[H_3O^+] = 8.9 \times 10^{-6}$
12. Calculate the pH Value
$pH = -log[H_3O^+]$
$pH = -log( 8.9 \times 10^{- 6})$
$pH = 5.05$
