Answer
(b) $pH = 2.8$
Work Step by Step
- 0.05 mole of $NaF$ in 1.0L = 0.05M NaF.
$[NaF] = [F^-] = 0.05M$ (Completely dissociated salt)
1. Drawing the ICE table, we get these concentrations at the equilibrium:
$HF(aq) + H_2O(l) \lt -- \gt F^-(aq) + H_3O^+(aq)$
Remember: Reactants at equilibrium = Initial Concentration - x
Products = Initial Concentration + x
$[HF] = 0.1 M - x$
$[F^-] = 0.05M + x$
$[H_3O^+] = 0 + x$
2. Calculate 'x' using the $K_a$ expression.
$ 7.1\times 10^{- 4} = \frac{[F^-][H_3O^+]}{[HF]}$
$ 7.1\times 10^{- 4} = \frac{( 0.05 + x )* x}{ 0.1 - x}$
Considering 'x' has a very small value.
$ 7.1\times 10^{- 4} = \frac{ 0.05 * x}{ 0.1}$
$ 7.1\times 10^{- 4} = 0.5x$
$\frac{ 7.1\times 10^{- 4}}{ 0.5} = x$
$x = 1.4\times 10^{- 3}$
Percent dissociation: $\frac{ 1.4\times 10^{- 3}}{ 0.1} \times 100\% = 1.4\%$
x = $[H_3O^+]$
3. Calculate the pH Value
$pH = -log[H_3O^+]$
$pH = -log( 1.4 \times 10^{- 3})$
$pH = 2.85$
