Answer
The pH of that buffer solution is equal to 5.26
Work Step by Step
$500$ mL = $500 \times 10^{-3}$ L = 0.500 L
1. Calculate the molar mass $(CH_3COONa)$:
12.01* 1 + 1.008* 3 + 12.01* 1 + 16.00*2 + 22.99* 1 = 82.03g/mol
2. Calculate the number of moles $(CH_3COONa)$
$n(moles) = \frac{mass(g)}{mm(g/mol)}$
$n(moles) = \frac{ 20.0}{ 82.03}$
$n(moles) = 0.244$
3. Find the concentration in mol/L $(CH_3COONa)$:
$C(mol/L) = \frac{n(moles)}{volume(L)}$
$ C(mol/L) = \frac{ 0.244}{ 0.500} $
$C(mol/L) = 0.488$
** Since $CH_3COONa$ is a strong electrolyte: $[CH_3COO^-]_{initial} = 0.488M$
4. 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
$[CH_3COOH] = 0.150 M - x$
$[CH_3COO^-] = 0.488M + x$
$[H_3O^+] = 0 + x$
5. 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{( 0.488 + x )* x}{ 0.15 - x}$
Considering 'x' has a very small value.
$ 1.8\times 10^{- 5} = \frac{ 0.488 * x}{ 0.150}$
$\frac{ 1.8\times 10^{- 5} * 0.150}{ 0.488} = x$
$x = 5.53\times 10^{- 6}$
Percent dissociation: $\frac{ 5.53\times 10^{- 6}}{ 0.150} \times 100\% = 3.69\times 10^{- 3}\%$
x = $[H_3O^+]$
6. Calculate the pH value
$pH = -log[H_3O^+]$
$pH = -log( 5.53 \times 10^{- 6})$
$pH = 5.26$
