Multiple Choice
A 100.0 mL sample of 0.10 M NH3 is titrated with 0.10 M HNO3. Determine the pH of the solution after the addition of 200.0 mL of HNO3. The Kb of NH3 is 1.8 × 10^-5.
18. Aqueous Equilibrium
Titrations: Weak Base-Strong Acid
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Consider the titration of 50.0 mL of 0.20 M NH3 (Kb=1.8×10⁻⁵) with 0.20 M HNO3. Calculate the pH after the addition of 50.0 mL of the titrant at 25 °C.
A
pH = 4.74
B
pH = 7.00
C
pH = 9.26
D
pH = 11.00
Verified step by step guidance1
Determine the initial moles of NH3 in the solution. Use the formula: \( \text{moles} = \text{concentration} \times \text{volume} \). Here, the concentration of NH3 is 0.20 M and the volume is 50.0 mL (convert to liters by dividing by 1000).
Calculate the moles of HNO3 added. Since the concentration of HNO3 is also 0.20 M and the volume added is 50.0 mL, use the same formula: \( \text{moles} = \text{concentration} \times \text{volume} \).
Determine the reaction between NH3 and HNO3. NH3 is a weak base and HNO3 is a strong acid. They react in a 1:1 molar ratio to form NH4+ and NO3-. Calculate the moles of NH3 and HNO3 remaining after the reaction.
Identify the resulting solution composition. After the reaction, you will have a solution of NH4+ (the conjugate acid of NH3) and possibly some excess NH3 or HNO3. Use the Henderson-Hasselbalch equation to find the pH: \( \text{pH} = \text{pK}_a + \log \left( \frac{[\text{base}]}{[\text{acid}]} \right) \). First, calculate \( \text{pK}_a \) from \( \text{pK}_w - \text{pK}_b \), where \( \text{pK}_w = 14 \).
Substitute the concentrations of NH4+ and NH3 into the Henderson-Hasselbalch equation to calculate the pH. Remember that the concentrations are based on the total volume of the solution after mixing, which is 100.0 mL (50.0 mL NH3 + 50.0 mL HNO3).
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