Page 56 - Analytical Chemistry I E-book
P. 56
• After addition of 20.00 mL of NaOH :
mmoles of acid left unreacted = (25 x 0.1) – (20 x 0.1) = 0.5 mmoles
volume of solution = 25 + 20 = 45 mL
[H+] = 0.5/45 = 0.0111 M pH = – log 0.0111 = 1.955
• After addition of 24.00 mL of NaOH :
mmoles of acid left unreacted = (25 x 0.1) – (24 x 0.1) = 0.1 mmoles
[H+] = 0.1 / (24 + 25) = 0.1/49 = 0.00204 M pH = 2.69
