Answer
(a) $pH = 2.87$
(b) $pH = 4.56$
(c) $pH = 5.34$
(d) $pH = 8.78$
(e) $pH = 12.10$
Work Step by Step
(a)
1. We have these concentrations at equilibrium:
-$[H_3O^+] = [CH_3COO^-] = x$
-$[CH_3COOH] = [CH_3COOH]_{initial} - x = 0.1 - x$
For approximation, we consider: $[CH_3COOH] = 0.1M$
2. Now, use the Ka value and equation to find the 'x' value.
$Ka = \frac{[H_3O^+][CH_3COO^-]}{ [CH_3COOH]}$
$Ka = 1.8 \times 10^{- 5}= \frac{x * x}{ 0.1}$
$Ka = 1.8 \times 10^{- 5}= \frac{x^2}{ 0.1}$
$ 1.8 \times 10^{- 6} = x^2$
$x = 1.3 \times 10^{- 3}$
Percent dissociation: $\frac{ 1.3 \times 10^{- 3}}{ 0.1} \times 100\% = 1.3\%$
%dissociation < 5% : Right approximation.
Therefore: $[H_3O^+] = [CH_3COO^-] = x = 1.3 \times 10^{- 3}M $
And, since 'x' has a very small value (compared to the initial concentration): $[CH_3COOH] \approx 0.1M$
3. Calculate the pH Value
$pH = -log[H_3O^+]$
$pH = -log( 1.3 \times 10^{- 3})$
$pH = 2.87$
(b)
1. Find the numbers of moles:
$C(CH_3COOH) * V(CH_3COOH) = 0.1* 0.025 = 2.5 \times 10^{-3}$ moles
$C(KOH) * V(KOH) = 0.2* 0.005 = 1 \times 10^{-3}$ moles
2. Write the acid-base reaction:
$CH_3COOH(aq) + KOH(aq) -- \gt KCH_3COO(aq) + H_2O(l)$
- Total volume: 0.025 + 0.005 = 0.03L
3. Since the base is the limiting reactant, only $ 0.001$ mol of the compounds will react.
Therefore:
Concentration (M) = $\frac{n(mol)}{Volume(L)}$
$[CH_3COOH] = 0.0025 - 0.001 = 1.5 \times 10^{-3}$ moles.
Concentration: $\frac{1.5 \times 10^{-3}}{ 0.03} = 0.05M$
$[KOH] = 0.001 - 0.001 = 0$
$[KCH_3COO] = 0 + 0.001 = 0.001$ moles.
Concentration: $\frac{ 0.001}{ 0.03} = 0.033M$
4. Calculate the $pK_a$ for the acid
$pKa = -log(Ka)$
$pKa = -log( 1.8 \times 10^{- 5})$
$pKa = 4.74$
5. Using the Henderson–Hasselbalch equation:
$pH = pKa + log(\frac{[Base]}{[Acid]})$
$pH = 4.74 + log(\frac{0.033}{0.05})$
$pH = 4.74 + log(0.67)$
$pH = 4.74 + -0.176$
$pH = 4.56$
(c)
1. Find the numbers of moles:
$C(CH_3COOH) * V(CH_3COOH) = 0.1* 0.025 = 2.5 \times 10^{-3}$ moles
$C(KOH) * V(KOH) = 0.2* 0.01 = 2 \times 10^{-3}$ moles
2. Write the acid-base reaction:
$CH_3COOH(aq) + KOH(aq) -- \gt KCH_3COO(aq) + H_2O(l)$
- Total volume: 0.025 + 0.01 = 0.035L
3. Since the base is the limiting reactant, only $ 0.002$ mol of the compounds will react.
Therefore:
Concentration (M) = $\frac{n(mol)}{Volume(L)}$
$[CH_3COOH] = 0.0025 - 0.002 = 5 \times 10^{-4}$ moles.
Concentration: $\frac{5 \times 10^{-4}}{ 0.035} = 0.014M$
$[KOH] = 0.002 - 0.002 = 0$
$[KCH_3COO] = 0 + 0.002 = 0.002$ moles.
Concentration: $\frac{ 0.002}{ 0.035} = 0.057M$
4. Calculate the $pK_a$ for the acid
$pKa = -log(Ka)$
$pKa = -log( 1.8 \times 10^{- 5})$
$pKa = 4.74$
5. Using the Henderson–Hasselbalch equation:
$pH = pKa + log(\frac{[Base]}{[Acid]})$
$pH = 4.74 + log(\frac{0.057}{0.014})$
$pH = 4.74 + log(4)$
$pH = 4.74 + 0.602$
$pH = 5.34$
(d)
1. Find the numbers of moles:
$C(CH_3COOH) * V(CH_3COOH) = 0.1* 0.025 = 2.5 \times 10^{-3}$ moles
$C(KOH) * V(KOH) = 0.2* 0.0125 = 2.5 \times 10^{-3}$ moles
2. When the number of moles is equal, the reactants are totally consumed:
$CH_3COOH(aq) + KOH(aq) -- \gt KCH_3COO(aq) + H_2O(l)$
- Total volume: 0.025 + 0.0125 = 0.0375L
3. So, those are the final concentrations:
$[CH_3COOH] = 0.0025 - 0.0025 = 0$ mol.
$[KOH] = 0.0025 - 0.0025 = 0$ mol
$[KCH_3COO] = 0 + 0.0025 = 0.0025$ moles.
Concentration: $\frac{ 0.0025}{ 0.0375} = 0.0666666666666667M$
- Therefore, we have a weak base salt solution:
- Since $CH_3COO^-$ is the conjugate base of $CH_3COOH$ , we can calculate its kb by using this equation:
$K_a * K_b = K_w = 10^{-14}$
$ 1.8\times 10^{- 5} * K_b = 10^{-14}$
$K_b = \frac{10^{-14}}{ 1.8\times 10^{- 5}}$
$K_b = 5.6\times 10^{- 10}$
4. We have these concentrations at equilibrium:
-$[OH^-] = [CH_3COOH] = x$
-$[CH_3COO^-] = [CH_3COO^-]_{initial} - x = 0.0666666666666667 - x$
For approximation, we consider: $[CH_3COO^-] = 0.0666666666666667M$
5. Now, use the Kb value and equation to find the 'x' value.
$Kb = \frac{[OH^-][CH_3COOH]}{ [CH_3COO^-]}$
$Kb = 5.6 \times 10^{- 10}= \frac{x * x}{ 0.067}$
$Kb = 5.6 \times 10^{- 10}= \frac{x^2}{ 0.067}$
$ 3.7 \times 10^{- 11} = x^2$
$x = 6.1 \times 10^{- 6}$
Percent ionization: $\frac{ 6.1 \times 10^{- 6}}{ 0.067} \times 100\% = 0.0091\%$
%ionization < 5% : Right approximation.
Therefore: $[OH^-] = [CH_3COOH] = x = 6.1 \times 10^{- 6}M $
$[CH_3COO^-] \approx 0.067M$
6. Calculate the pOH:
$pOH = -log[OH^-]$
$pOH = -log( 6.1 \times 10^{- 6})$
$pOH = 5.22$
7. Find the pH:
$pH + pOH = 14$
$pH + 5.22 = 14$
$pH = 8.78$
(e)
1. Find the numbers of moles:
$C(CH_3COOH) * V(CH_3COOH) = 0.1* 0.025 = 2.5 \times 10^{-3}$ moles
$C(KOH) * V(KOH) = 0.2* 0.015 = 3 \times 10^{-3}$ moles
Write the acid-base reaction:
$CH_3COOH(aq) + KOH(aq) -- \gt KCH_3COO(aq) + H_2O(l)$
- Total volume: 0.025 + 0.015 = 0.04L
Since the acid is the limiting reactant, only $ 0.0025$ mol of the compounds will react.
Therefore:
Concentration (M) = $\frac{n(mol)}{Volume(L)}$
$[CH_3COOH] = 0.0025 - 0.0025 = 0M$.
$[KOH] = 0.003 - 0.0025 = 5 \times 10^{-4}$ mol
Concentration: $\frac{5 \times 10^{-4}}{ 0.04} = 0.012M$
$[KCH_3COO] = 0 + 0.0025 = 0.0025$ moles.
Concentration: $\frac{ 0.0025}{ 0.04} = 0.063M$
- We have a strong and a weak base. We can ignore the weak one and calculate the pH based only on the strong base concentration:
$[OH^-] = [KOH]$
2. Calculate the pOH:
$pOH = -log[OH^-]$
$pOH = -log( 0.012)$
$pOH = 1.90$
3. Find the pH:
$pH + pOH = 14$
$pH + 1.90 = 14$
$pH = 12.10$
