Page 57 - Analytical Chemistry I E-book
P. 57
• After addition of 24.50 mL of NaOH :
mmoles of acid left unreacted = (25 x 0.1) – (24.5 x 0.1) = 0.05 mmoles
[H+] = 0.05/49.50 = 0.00101 M pH = 3.00
• After addition of 24.90 mL of NaOH :
mmoles of acid left unreacted = (25 x 0.1) – (24.9 x 0.1) = 0.01 mmoles
[H+] = 0.01/49.90 = 0.00020 M pH = 3.70
• After addition of 25.00 mL of NaOH : (The equivalence point)
No acid or base left unreacted; the solution contains only NaCl which is a salt
of strong acid and strong base. pH = 7.00
There is a jump of more than 3 pH units due to addition of 0.1 mL of
NaOH.
