Answer
The pH of that solution is equal to 3.82
Work Step by Step
1000ml = 1L
55ml = 0.055 L
125ml = 0.125 L
1. Find the numbers of moles of each compound:
$C(HF) * V(HF) = 0.050*0.055 = 2.75 \times 10^{-3}$ moles
$C(NaF) * V(NaF) = 0.10*0.125 = 0.0125$ moles
2. Calculate the total volume:
- Total volume: 0.055 + 0.125 = 0.180L
3. Calculate the final concentrations:
$[HF] : \frac{2.75 \times 10^{-3}}{0.180} = 0.0153M $
$[NaF] : \frac{0.0125}{0.180} = 0.0694M $
- Since $NaF$ is a strong electrolyte: $[NaF] = [F^-] = 0.0694M$
4. 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
And Products = Initial Concentration + x
$[HF] = 0.0153 M - x$
$[F^-] = 0.0694M + x$
$[H_3O^+] = 0 + x$
5. Calculate 'x' using the $K_a$ expression.
$ 6.8\times 10^{- 4} = \frac{[F^-][H_3O^+]}{[HF]}$
$ 6.8\times 10^{- 4} = \frac{( 0.0694 + x )* x}{ 0.0153 - x}$
Considering 'x' has a very small value.
$ 6.8\times 10^{- 4} = \frac{ 0.0694 * x}{ 0.0153}$
$ 6.8\times 10^{- 4} = 4.54x$
$\frac{ 6.8\times 10^{- 4}}{ 4.54} = x$
$x = 1.5\times 10^{- 4}$
Percent dissociation: $\frac{ 1.5\times 10^{- 4}}{ 0.015} \times 100\% = 0.98\%$
x = $[H_3O^+]$
6. Calculate the pH value
$pH = -log[H_3O^+]$
$pH = -log( 1.5 \times 10^{- 4})$
$pH = 3.82$
