Question

Calculate the pH for each of the following cases in the titration of 35.0 mL of 0.120 M NaOH(aq), with 0.120 M HCl(aq).

(a) before addition of any HCl

(b) after addition of 13.5 mL of HCl

(c) after addition of 21.5 mL of HCl

(d) after the addition of 35.0 mL of HCl

(e) after the addition of 43.5 mL of HCl

(f) after the addition of 50.0 mL of HCl

Answer 1

millimoles of NaOH = 0.120 x 35 = 4.2

a) before addition of any HCl

NaOH = 0.120 M

[OH-] = 0.120 M

pOH = -log[OH-] = -log (0.120) = 0.92

pH +pOH = 14

pH = 13.08

b) 13.5 ml HCl 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)

              = 0.053 M

pOH = 1.27

pH = 12.73

c) 21.5 ml HCl added

millimoles of HI = 21.5 x 0.120 = 2.58

millimoles of base > millimoles of acid

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

               =0.0361

           pOH = 1.44

pH = 12.56

d) 35 ml HI added

millimoles of base = millimoles of acid

so it equivalence point . strong acid + strong base

pH = 7.00

e) 40.5 ml HI added

millimoles of HI = 43.5 x 0.120 = 5.22

millimoles of base < millimoles of acid

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

               =0.0665 M

pH = -log[H+]

pH = 1.18

f) 50 ml HI added

millimoles of HI = 50 x 0.120 =6.0

millimoles of base < millimoles of acid

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

              

            = 0.0212 M

pH = -log[H+]

pH = 1.67

millimoles of NaOH = 0.170 x 35 = 5.95