Multiple Choice
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.
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18. Aqueous Equilibrium
Titrations: Weak Base-Strong Acid
Multiple Choice
What is the approximate pH at the equivalence point of a weak base-strong acid titration if 50.00 mL of aqueous ammonia (NH3) requires 37.10 mL of 0.1087 mol L^-1 HCl? pKb = 4.75 for ammonia. Report your answer to 2 decimal places.
A
4.75
B
5.25
C
7.00
D
9.25
Verified step by step guidance1
First, understand that at the equivalence point of a weak base-strong acid titration, the moles of acid added are equal to the moles of base present initially. Calculate the moles of HCl added using the formula: \( \text{moles of HCl} = \text{volume of HCl} \times \text{concentration of HCl} \).
Next, calculate the initial moles of ammonia (NH3) present using the formula: \( \text{moles of NH3} = \text{volume of NH3} \times \text{concentration of NH3} \). Since the moles of HCl added are equal to the moles of NH3 initially present, you can use this to find the concentration of the ammonium ion (NH4^+) formed at the equivalence point.
At the equivalence point, the solution contains ammonium ions (NH4^+), which is the conjugate acid of ammonia. Use the formula for the equilibrium constant of the conjugate acid, \( K_a = \frac{K_w}{K_b} \), where \( K_w \) is the ion-product constant of water (\( 1.0 \times 10^{-14} \) at 25°C) and \( K_b \) is given as 4.75.
Set up the expression for the dissociation of NH4^+ in water: \( NH4^+ \rightarrow NH3 + H^+ \). Use the expression for \( K_a \) to find \( [H^+] \) at equilibrium: \( K_a = \frac{[NH3][H^+]}{[NH4^+]} \). Assume \( [NH3] \approx [H^+] \) and \( [NH4^+] \) is the initial concentration of NH4^+.
Finally, calculate the pH using the formula: \( \text{pH} = -\log[H^+] \). This will give you the approximate pH at the equivalence point.
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