Question

Part A For 550.0 mL of pure water, calculate the initial pH and the final pH after adding 0.010 mol of HCl .

Part B For 550.0 mL of a buffer solution that is 0.155 M in HC 2 H 3 O 2 and 0.140 M in NaC 2 H 3 O 2 , calculate the initial pH and the final pH after adding 0.010 mol of HCl .

Part C For 550.0 mL of a buffer solution that is 0.150 M in CH 3 CH 2 NH 2 and 0.135 M in CH 3 CH 2 NH 3 Cl , calculate the initial pH and the final pH after adding 0.010 mol of HCl

Answer 1

Part A for pure water

initial pH (no HCl) = 7

Final pH with HCl = -log[H+] = -log(0.01/0.55) = 1.70

Part BFor buffer

using Hendersen-Hasselbalck equation,

pH = pKa + log(base/acid)

initial pH = 4.75 + log(0.14/0.155) = 4.79

After 0.01 mol Hcl added

final pH = 4.75 + log[(0.14 x 0.55 - 0.01)/(0.155 x 0.55 + 0.01)] = 4.903

Part C For buffer

initial pH = 10.63 - log(0.15/0.135) = 10.584

after 0.01 mol HCl added

final pH = 10.63 + log[(0.15 x 0.55 - 0.01)/(0.135 x 0.55 + 0.01)] = 10.695

Part C