Question

Construct a curve for the titration of 50.00 mL of a 0.1000 M solution of compound A with a 0.2000 M solution of compound B in the following table. For each titration, calculate the pH after the addition of 0.00, 12.50, 20.00, 24.00, 25.00, 26.00, 37.50, 45.00, 49.00, 50.00, 51.00, and 60.00 mL of com- pound B.

A

(a) H2SO3

(b) ethylenediamine

(c) H2SO4

B

NaOH

HCl

NaOH

Answer 1

titration between H2SO4 (from A) and NaOH from (B)

millimoles of H2SO4 = 2 x 50 x 0.1 = 10    (here 2 indicates 2 H+)

1) 0.00 ml NaOH added

H2SO4 molarity = 0.1 x 2 = 0.2

pH = -log [H+] = -log 0.2 = 0.699

2) 20 ml NaOH added

millimoles of NaOH = 20 x 0.2 = 4

millimoles of H2SO4 > millimoles of NaOH

so acid dominent

[H+] = (millimoles of H2SO4 - millimoles of NaOH) / total volume

       = (10 - 4) / (50 +20)

      = 0.0857 M

pH = 1.067

3) 24 ml NaOH added

millimoles of NaOH = 24 x 0.2 = 4.8

millimoles of H2SO4 > millimoles of NaOH

so acid dominent

[H+] = (millimoles of H2SO4 - millimoles of NaOH) / total volume

       = (10 - 4.8) / (50 +24)

      = 0.0703 M

pH = 1.153

4) 25 ml NaOH added

millimoles of NaOH = 25 x 0.2 = 5

millimoles of H2SO4 > millimoles of NaOH

so acid dominent

[H+] = (millimoles of H2SO4 - millimoles of NaOH) / total volume

       = (10 - 5) / (50 +25)

      = 0.0667M

pH = 1.176

5) 26 ml NaOH added

millimoles of NaOH = 26 x 0.2 = 5.2

millimoles of H2SO4 > millimoles of NaOH

so acid dominent

[H+] = (millimoles of H2SO4 - millimoles of NaOH) / total volume

       = (10 - 5.2) / (50 +26)

      = 0.063 M

pH = 1.20

6) 37.5 ml NaOH added

millimoles of NaOH = 37.5 x 0.2 = 7.5

millimoles of H2SO4 > millimoles of NaOH

so acid dominent

[H+] = (millimoles of H2SO4 - millimoles of NaOH) / total volume

       = (10 - 7.5) / (50 +37.5)

      = 0.02857 M

pH = 1.544

7) 20 ml NaOH added

millimoles of NaOH = 45 x 0.2 = 9

millimoles of H2SO4 > millimoles of NaOH

so acid dominent

[H+] = (millimoles of H2SO4 - millimoles of NaOH) / total volume

       = (10 - 9) / (50 +45)

      = 0.010 M

pH = 1.977

8) 49 ml NaOH added

millimoles of NaOH = 49 x 0.2 = 9.8

millimoles of H2SO4 > millimoles of NaOH

so acid dominent

[H+] = (millimoles of H2SO4 - millimoles of NaOH) / total volume

       = (10 - 9.8) / (50 +49)

      = 2.02 x 10^-3 M

pH = 2.69

9) 50 ml NaOH added

millimoles of NaOH = 50 x 0.2 = 10

millimoles of acid = millimoles of base

so it is equivalece point . strong acid + strong base here pH= 7

10) 60 ml NaOH added

millimoles of NaOH = 60 x 0.2 = 12

millimoles of H2SO4 < millimoles of NaOH

so acid dominent

[OH] = (millimoles of NaOH - millimoles of H2SO4) / total volume

       = (12 - 10) / (50 +60)

      = 0.0182 M

pOH = -log [OH-]

pOH = 1.74

pH + pOH = 14

pH = 12.26

note : you see the length of the problem . for one combination we get 10 pH's for all combination just imagine this problem length. remaining combinations try yourself.