Multiple Choice
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.
18. Aqueous Equilibrium
Titrations: Weak Base-Strong Acid
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Consider the titration of a 26.0-mL sample of 0.175 M CH3NH2 (Kb = 4.4 × 10^(-4)) with 0.150 M HBr. Determine the pH at one-half of the equivalence point.
A
pH = 10.63
B
pH = 9.24
C
pH = 5.76
D
pH = 7.00
Verified step by step guidance1
Identify that CH3NH2 is a weak base and HBr is a strong acid. The titration involves the reaction: CH3NH2 + HBr -> CH3NH3+ + Br-.
At one-half of the equivalence point, half of the CH3NH2 has been converted to CH3NH3+. This means the concentrations of CH3NH2 and CH3NH3+ are equal.
Use the Henderson-Hasselbalch equation for a base: pH = pKa + log([base]/[acid]). Here, [base] = [CH3NH2] and [acid] = [CH3NH3+].
Calculate pKa from Kb using the relation: pKa + pKb = 14. First, find pKb: pKb = -log(4.4 × 10^(-4)). Then, calculate pKa.
Since [CH3NH2] = [CH3NH3+] at one-half equivalence point, the log term in the Henderson-Hasselbalch equation becomes zero, simplifying the equation to pH = pKa.
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