Question

Calculate the pH for each of the following cases in the titration of 35.0 mL of 0.120 M LiOH(aq), with 0.120 M HI(aq). (a) before addition of any HI (b) after addition of 13.5 mL of HI (c) after addition of 20.5 mL of HI (d) after the addition of 35.0 mL of HI (e) after the addition of 45.5 mL of HI (f) after the addition of 50.0 mL of HI

Answer 1

millimoles of LiOH = 0.120 x 35 = 4.2

a) before addition of any HI

LiOH = 0.120 M

[OH-] = 0.120 M

pOH = -log[OH-] = -log (0.120) = 0.921

pH +pOH = 14

pH = 13.08

b) 13.5 ml HI added

millimoles of HI = 13.5 x 0.120 = 1.62

millimoles of base > millimoles of acid

so [OH-] = (base millimoles - acid millimoles )/ total volume

               = (4.2 - 1.62) / (35+13.5)

              = 0.0532M

pOH = 1.27

pH = 12.73

c) 20.5 ml HI added

millimoles of HI = 20.5 x 0.120 = 2.46

millimoles of base > millimoles of acid

so [OH-] = (base millimoles - acid millimoles )/ total volume

               = (4.2 - 2.46) / (35+20.5)

              = 0.0314M

pOH = 1.50

pH = 12.50

d) 35 ml HI added

millimoles of HI = 35 x 0.120 = 4.2

millimoles of base = millimoles of acid

so it equivalence point . strong acid + strong base

pH = 7.00