Answer
Initial buffer: $pH = 3.76$
(a) $pH = 3.71$
(b) $pH = 3.87$
Work Step by Step
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
And Products = Initial Concentration + x
$[HF] = 0.25 M - x$
$[F^-] = 0.5M + x$
$[H_3O^+] = 0 + x$
2. Calculate 'x' using the $K_a$ expression.
$ 3.5\times 10^{- 4} = \frac{[F^-][H_3O^+]}{[HF]}$
$ 3.5\times 10^{- 4} = \frac{( 0.5 + x )* x}{ 0.25 - x}$
Considering 'x' has a very small value.
$ 3.5\times 10^{- 4} = \frac{ 0.5 * x}{ 0.25}$
$ 3.5\times 10^{- 4} = 2x$
$\frac{ 3.5\times 10^{- 4}}{ 2} = x$
$x = 1.75\times 10^{- 4}$
Percent dissociation: $\frac{ 1.75\times 10^{- 4}}{ 0.25} \times 100\% = 0.07\%$
x = $[H_3O^+]$
$[HF] = 0.25 M - x = 0.25 M - 1.75 \times 10^{-4}M \approx 0.25M$
$[F^-] = 0.5M + x = 0.25 M + 1.75 \times 10^{-4}M \approx 0.50M$
$[H_3O^+] = 0 + x = 1.75 \times 10^{-4}M$
3. Calculate the pH Value
$pH = -log[H_3O^+]$
$pH = -log( 1.75 \times 10^{- 4})$
$pH = 3.757 \approx 3.76$
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$(a)$
4. Since we are adding a strong acid, this reaction will occur:
$F^-(aq) + H_3O^+(aq) -- \gt HF(aq) + H_2O(l)$
And these are the concentrations after this reaction:
Remember: Reactants at equilibrium = Initial Concentration - y
And Products = Initial Concentration + y
Concentration (Added acid) = $\frac{n(moles)}{volume(L)} = \frac{ 0.002}{ 0.1} = 0.02$
Since $HNO_3$ is a strong base, y = $[HNO_3] = 0.02M$
$[HF] = 0.25 M + 0.02 = 0.27M$
$[F^-] = 0.5M - 0.02 = 0.48M$
5. Now, calculate the hydronium ion concentration after the addition of the $HNO_3$:
$[H_3O^+] = Ka * (\frac{[HF]}{[F^-]})$
$[H_3O^+] = 3.5 \times 10^{-4} * \frac{0.27}{0.48}$
$[H_3O^+] = 3.5 \times 10^{-4} * 0.563$
$[H_3O^+] = 1.97 \times 10^{-4}$
6. Calculate the pH Value
$pH = -log[H_3O^+]$
$pH = -log( 1.97 \times 10^{- 4})$
$pH = 3.706 \approx 3.71$
$(b)$
4. Since we are adding a strong base, this reaction will occur:
$HF(aq) + OH^-(aq) -- \gt F^-(aq) + H_2O(l)$
And these are the concentrations after this reaction:
Remember: Reactants at equilibrium = Initial Concentration - y
And Products = Initial Concentration + y
Concentration (Added base) = $\frac{n(moles)}{volume(L)} = \frac{ 0.004}{ 0.1} = 0.04$
Since $KOH$ is a strong base, y = $[KOH] = 0.04M$
$[HF] = 0.25 M - 0.04 = 0.21M$
$[F^-] = 0.5M + 0.04 = 0.54M$
5. Now, calculate the hydronium ion concentration after the addition of the $KOH$:
$[H_3O^+] = Ka * (\frac{[HF]}{[F^-]})$
$[H_3O^+] = 3.5 \times 10^{-4} * \frac{0.21}{0.54}$
$[H_3O^+] = 3.5 \times 10^{-4} * 0.389$
$[H_3O^+] = 1.36 \times 10^{-4}$
6. Calculate the pH Value
$pH = -log[H_3O^+]$
$pH = -log( 1.36 \times 10^{- 4})$
$pH = 3.866 \approx 3.87$
