Answer :
To find the final pH when 200.0 mL of a [tex]\(0.200 \text{ M }\text{NH}_4\text{Cl (aq)}\)[/tex] is mixed with 100.0 mL of a [tex]\(0.200 \text{ M NaOH (aq)}\)[/tex], we need to follow these detailed steps:
1. Calculate the moles of each reactant:
- For [tex]\(\text{NH}_4\text{Cl}\)[/tex]:
[tex]\[ \text{moles NH}_4\text{Cl} = \text{Volume (L)} \times \text{Molarity (M)} = 0.200 \, \text{L} \times 0.200 \, \text{M} = 0.040 \, \text{moles} \][/tex]
- For [tex]\(\text{NaOH}\)[/tex]:
[tex]\[ \text{moles NaOH} = \text{Volume (L)} \times \text{Molarity (M)} = 0.100 \, \text{L} \times 0.200 \, \text{M} = 0.020 \, \text{moles} \][/tex]
2. Determine the reaction that takes place:
[tex]\[ \text{NH}_4\text{Cl} \rightarrow \text{NH}_4^+ + \text{Cl}^- \][/tex]
[tex]\[ \text{NaOH} \rightarrow \text{Na}^+ + \text{OH}^- \][/tex]
[tex]\[ \text{NH}_4^+ + \text{OH}^- \rightarrow \text{NH}_3 + \text{H}_2\text{O} \][/tex]
3. Identify the limiting reagent:
- [tex]\(\text{NH}_4^+\)[/tex] reacts with [tex]\(\text{OH}^-\)[/tex].
- Compare the moles: [tex]\(0.040 \, \text{moles NH}_4^+\)[/tex] and [tex]\(0.020 \, \text{moles OH}^-\)[/tex].
Therefore, [tex]\(\text{OH}^-\)[/tex] is the limiting reagent.
4. Calculate the remaining moles after the reaction:
- [tex]\(\text{OH}^-\)[/tex] is completely consumed, leaving 0 moles of [tex]\(\text{OH}^-\)[/tex].
- [tex]\(\text{NH}_4^+\)[/tex] remaining:
[tex]\[ \text{remaining moles NH}_4^+ = 0.040 \, \text{moles} - 0.020 \, \text{moles} = 0.020 \, \text{moles} \][/tex]
- [tex]\(\text{NH}_3\)[/tex] produced:
[tex]\[ \text{moles NH}_3 = 0.020 \, \text{moles} \][/tex]
5. Calculate total volume of the solution:
[tex]\[ \text{Total volume} = 200.0 \, \text{mL} + 100.0 \, \text{mL} = 300.0 \, \text{mL} = 0.300 \, \text{L} \][/tex]
6. Calculate concentrations in the resulting solution:
- [tex]\(\text{NH}_4^+\)[/tex]:
[tex]\[ \text{[NH}_4^+\text{]} = \frac{\text{moles NH}_4^+}{\text{total volume}} = \frac{0.020 \, \text{moles}}{0.300 \, \text{L}} = 0.067 \, \text{M} \][/tex]
- [tex]\(\text{NH}_3\)[/tex]:
[tex]\[ \text{[NH}_3\text{]} = \frac{\text{moles NH}_3}{\text{total volume}} = \frac{0.020 \, \text{moles}}{0.300 \, \text{L}} = 0.067 \, \text{M} \][/tex]
7. Set up an expression for [tex]\([OH^-]\)[/tex] using the [tex]\(\text{Kb}\)[/tex] for [tex]\(\text{NH}_3\)[/tex]:
The equilibrium for [tex]\(\text{NH}_3\)[/tex] in water:
[tex]\[ \text{NH}_3 + \text{H}_2\text{O} \leftrightharpoons \text{NH}_4^+ + \text{OH}^- \][/tex]
For this, we use the [tex]\(K_b\)[/tex] expression:
[tex]\[ K_b = \frac{[\text{NH}_4^+][\text{OH}^-]}{[\text{NH}_3]} \][/tex]
Given [tex]\(K_b(\text{NH}_3) = 1.80 \times 10^{-5}\)[/tex]:
[tex]\[ 1.80 \times 10^{-5} = \frac{(0.067 + x)x}{0.067} \][/tex]
Solving for [tex]\(x\)[/tex]:
[tex]\[ 1.80 \times 10^{-5} = \frac{0.067x + x^2}{0.067} \][/tex]
Assuming [tex]\(x\)[/tex] is small relative to initial concentration of [tex]\(\text{NH}_4^+\)[/tex] (0.067 M):
[tex]\[ 1.80 \times 10^{-5} \approx \frac{0.067x}{0.067} \][/tex]
[tex]\[ x \approx 1.80 \times 10^{-5} \][/tex]
Hence, [tex]\([\text{OH}^-] \approx 1.80 \times 10^{-5} \, \text{M}\)[/tex].
8. Calculate [tex]\(pOH\)[/tex] and then [tex]\(pH\)[/tex]:
[tex]\[ pOH = -\log_{10}([\text{OH}^-]) = -\log_{10}(1.80 \times 10^{-5}) \approx 4.74 \][/tex]
[tex]\[ pH = 14 - pOH = 14 - 4.74 = 9.26 \][/tex]
Final pH: The final pH of the solution is approximately [tex]\(9.26\)[/tex].
1. Calculate the moles of each reactant:
- For [tex]\(\text{NH}_4\text{Cl}\)[/tex]:
[tex]\[ \text{moles NH}_4\text{Cl} = \text{Volume (L)} \times \text{Molarity (M)} = 0.200 \, \text{L} \times 0.200 \, \text{M} = 0.040 \, \text{moles} \][/tex]
- For [tex]\(\text{NaOH}\)[/tex]:
[tex]\[ \text{moles NaOH} = \text{Volume (L)} \times \text{Molarity (M)} = 0.100 \, \text{L} \times 0.200 \, \text{M} = 0.020 \, \text{moles} \][/tex]
2. Determine the reaction that takes place:
[tex]\[ \text{NH}_4\text{Cl} \rightarrow \text{NH}_4^+ + \text{Cl}^- \][/tex]
[tex]\[ \text{NaOH} \rightarrow \text{Na}^+ + \text{OH}^- \][/tex]
[tex]\[ \text{NH}_4^+ + \text{OH}^- \rightarrow \text{NH}_3 + \text{H}_2\text{O} \][/tex]
3. Identify the limiting reagent:
- [tex]\(\text{NH}_4^+\)[/tex] reacts with [tex]\(\text{OH}^-\)[/tex].
- Compare the moles: [tex]\(0.040 \, \text{moles NH}_4^+\)[/tex] and [tex]\(0.020 \, \text{moles OH}^-\)[/tex].
Therefore, [tex]\(\text{OH}^-\)[/tex] is the limiting reagent.
4. Calculate the remaining moles after the reaction:
- [tex]\(\text{OH}^-\)[/tex] is completely consumed, leaving 0 moles of [tex]\(\text{OH}^-\)[/tex].
- [tex]\(\text{NH}_4^+\)[/tex] remaining:
[tex]\[ \text{remaining moles NH}_4^+ = 0.040 \, \text{moles} - 0.020 \, \text{moles} = 0.020 \, \text{moles} \][/tex]
- [tex]\(\text{NH}_3\)[/tex] produced:
[tex]\[ \text{moles NH}_3 = 0.020 \, \text{moles} \][/tex]
5. Calculate total volume of the solution:
[tex]\[ \text{Total volume} = 200.0 \, \text{mL} + 100.0 \, \text{mL} = 300.0 \, \text{mL} = 0.300 \, \text{L} \][/tex]
6. Calculate concentrations in the resulting solution:
- [tex]\(\text{NH}_4^+\)[/tex]:
[tex]\[ \text{[NH}_4^+\text{]} = \frac{\text{moles NH}_4^+}{\text{total volume}} = \frac{0.020 \, \text{moles}}{0.300 \, \text{L}} = 0.067 \, \text{M} \][/tex]
- [tex]\(\text{NH}_3\)[/tex]:
[tex]\[ \text{[NH}_3\text{]} = \frac{\text{moles NH}_3}{\text{total volume}} = \frac{0.020 \, \text{moles}}{0.300 \, \text{L}} = 0.067 \, \text{M} \][/tex]
7. Set up an expression for [tex]\([OH^-]\)[/tex] using the [tex]\(\text{Kb}\)[/tex] for [tex]\(\text{NH}_3\)[/tex]:
The equilibrium for [tex]\(\text{NH}_3\)[/tex] in water:
[tex]\[ \text{NH}_3 + \text{H}_2\text{O} \leftrightharpoons \text{NH}_4^+ + \text{OH}^- \][/tex]
For this, we use the [tex]\(K_b\)[/tex] expression:
[tex]\[ K_b = \frac{[\text{NH}_4^+][\text{OH}^-]}{[\text{NH}_3]} \][/tex]
Given [tex]\(K_b(\text{NH}_3) = 1.80 \times 10^{-5}\)[/tex]:
[tex]\[ 1.80 \times 10^{-5} = \frac{(0.067 + x)x}{0.067} \][/tex]
Solving for [tex]\(x\)[/tex]:
[tex]\[ 1.80 \times 10^{-5} = \frac{0.067x + x^2}{0.067} \][/tex]
Assuming [tex]\(x\)[/tex] is small relative to initial concentration of [tex]\(\text{NH}_4^+\)[/tex] (0.067 M):
[tex]\[ 1.80 \times 10^{-5} \approx \frac{0.067x}{0.067} \][/tex]
[tex]\[ x \approx 1.80 \times 10^{-5} \][/tex]
Hence, [tex]\([\text{OH}^-] \approx 1.80 \times 10^{-5} \, \text{M}\)[/tex].
8. Calculate [tex]\(pOH\)[/tex] and then [tex]\(pH\)[/tex]:
[tex]\[ pOH = -\log_{10}([\text{OH}^-]) = -\log_{10}(1.80 \times 10^{-5}) \approx 4.74 \][/tex]
[tex]\[ pH = 14 - pOH = 14 - 4.74 = 9.26 \][/tex]
Final pH: The final pH of the solution is approximately [tex]\(9.26\)[/tex].
