Multiple Choice
A 20.00-mL sample of 0.150 M NH3 is being titrated with 0.200 M HCl. What is the pH after 25.00 mL of HCl has been added? Kb of NH3 = 1.8 × 10⁻⁵
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18. Aqueous Equilibrium
Titrations: Weak Base-Strong Acid
Multiple Choice
Consider the titration of a 26.0-mL sample of 0.185 M CH3NH2 (Kb=4.4×10⁻⁴) with 0.155 M HBr. What is the pH after adding 6.0 mL of HBr?
A
4.75
B
5.28
C
6.02
D
7.00
Verified step by step guidance1
Calculate the initial moles of CH3NH2 using the formula: moles = concentration × volume. Convert the volume from mL to L before calculating.
Calculate the moles of HBr added using the same formula: moles = concentration × volume. Again, ensure the volume is in liters.
Determine the moles of CH3NH2 and HBr that react. Since HBr is a strong acid and CH3NH2 is a weak base, they will react in a 1:1 ratio to form CH3NH3+ and Br-. Subtract the moles of HBr from the initial moles of CH3NH2 to find the moles of CH3NH2 remaining.
Calculate the concentration of CH3NH3+ formed and the remaining CH3NH2 in the solution. Use the total volume of the solution (initial volume of CH3NH2 + volume of HBr added) to find these concentrations.
Use the Henderson-Hasselbalch equation to find the pH of the solution: \( \text{pH} = \text{pK}_a + \log \left( \frac{[\text{base}]}{[\text{acid}]} \right) \). First, calculate \( \text{pK}_a \) from \( \text{pK}_b \) using the relation \( \text{pK}_a + \text{pK}_b = 14 \). Then, substitute the concentrations of CH3NH2 (base) and CH3NH3+ (acid) into the equation to find the pH.
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