Answer
$pH = 3.85$
Work Step by Step
1. Calculate the molar mass $(NaCH3CO2)$:
22.99* 1 + 12.01* 1 + 1.01* 3 + 12.01* 1 + 16* 2 = 82.04g/mol
2. Calculate the number of moles $(NaCH3CO2)$
$n(moles) = \frac{mass(g)}{mm(g/mol)}$
$n(moles) = \frac{ 1.56}{ 82.04}$
$n(moles) = 0.0190$
3. Find the concentration in mol/L $(NaCH3CO2)$:
$0.019$ mol in 1L: $0.019 M (NaCH3CO2)$
4. 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
$[CH_3CO_2H] = 0.15 M - x$
$[CH_3C{O_2}^-] = 0.019M + x$
$[H_3O^+] = 0 + x$
5. 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{( 0.019 + x )* x}{ 0.15 - x}$
Considering 'x' has a very small value.
$ 1.8\times 10^{- 5} = \frac{ 0.019 * x}{ 0.15}$
$ 1.8\times 10^{- 5} = 0.13x$
$\frac{ 1.8\times 10^{- 5}}{ 0.13} = x$
$x = 1.4\times 10^{- 4}$
Percent dissociation: $\frac{ 1.4\times 10^{- 4}}{ 0.15} \times 100\% = 0.095\%$
x = $[H_3O^+]$
6. Calculate the pH Value
$pH = -log[H_3O^+]$
$pH = -log( 1.4 \times 10^{- 4})$
$pH = 3.85$
