Page 58 - Analytical Chemistry I E-book
P. 58
• After addition of 25.10 mL of NaOH :
mmoles of base added = 25.10 x 0.1 = 2.51
mmoles of acid reacted (mmoles acid at start) = 2.50
mmoles of base unreacted = 2.51 – 2.50 = 0.01
[OH–] = 0.01/50.1 = 0.0001996 M
pOH = 3.70 pH = 14 – 3.70 = 10.30
• There is a jump of about 7 pH units (3.70 ⎯→ 10.30) due to the addition of
only 0.2 mL of the titrant (from 24.9 to 25.1 mL).
