Question

(a) How many mL of a 0.1 N solution of NaOH must be added to a 25 mL solution of 0.1 N Acetic Acid to obtain a pH of 9.75? What is the concentration of Acetic Acid at this pH?

(b) Answer the same question with the following changes: [Acetic Acid] = 0.01 N in25mL and the normality of the titrant is 0.02 N. The final pH is 11.

(c) What are the alkalinity and acidity of a water that contains 79 mg/L of HCO₃^- and has a pH of 7.3? Assume a temperature of 25^oC.

Answer 1

(a) Volume of a 0.1 N solution of NaOH must be added to a 25 mL solution of 0.1 N Acetic Acid to obtain a pH of 9.75 can be calculated as follows:

N(NaOH) V(NaOH) = N(CH3COOH) V(CH3COOH)

0.1 x V(NaOH) = 0.1 x 25

V(NaOH) = 25 mL

concentration of Acetic Acid at this pH = 9.75, we will use Henderson-Hasselbalch equation as follows

pH = pKa + log ( [A-] / [HA] )...............(1)

pH = 9.75 , pka = - log Ka = - log (1.8 x 10^-5) = 4.744

CH3COOH <=> CH3COO- + H+

Initial 0.1 0 0

Equilibrium 0.1 +x +x

final 0.1 - x x x

Ka = x^2 / (0.1 - x)

  1.8*10-5 = x^2

x =   1.8*10-5 = 1.34 x 10^-5 N or M as x factor = 1 for acetic acid

So [A-] = x = 1.34 x 10^-5 N or M

Substitute all in equation (1)

pH = pKa + log ( [A-] / [HA] )

9.75 = 4.744 + log ( 1.34 x 10^-5 / [HA] )

log ( 1.34 x 10^-5 / [HA] ) = 9.75 - 4.744 = 5.006

1.34 x 10^-5 / [HA] = antilog(5.006)

1.34 x 10^-5 / [HA] = 101391.139

[HA] = 1.34 x 10^-5 / 101391.139 = 1.32 x 10^-10 N

(b) [Acetic Acid] = 0.01 N in 25mL and the normality of the titrant is 0.02 N. The final pH is 11

  N(NaOH) V(NaOH) = N(CH3COOH) V(CH3COOH)

0.02 x V(NaOH) = 0.01 N x 25mL

V(NaOH) = (0.01 N x 25mL) / 0.02 = 12.5 mL

Ka = x^2 / (0.01 - x)

1.8*10-5 = x^2 [ (0.01 - x) can be neglected as above )

x =   1.8*10-5 = 1.34 x 10^-5 N or M as x factor = 1 for acetic acid

Substitute all in equation (1)

pH = pKa + log ( [A-] / [HA] )

11 = 4.744 + log ( 1.34 x 10^-5 / [HA] )

log ( 1.34 x 10^-5 / [HA] ) = 11 - 4.744 = 6.256

1.34 x 10^-5 / [HA] = antilog(6.256)

1.34 x 10^-5 / [HA] = 1803017.74

[HA] = 1.34 x 10^-5 / 1803017.74 = 7.43 x 10^-12 N

(c) The alkalinity and acidity of a water that contains 79 mg/L of HCO3- and has a pH of 7.3 at Temperature of 25 degree celcius.

H₂CO₃   ⇄    H⁺ + HCO₃^-

Ka = [H+] [HCO₃^- ] / [H₂CO₃ ]

4.3 x 10-7 = 79^2 / [H₂CO₃ ]

[H₂CO₃ ] = 1.45 M = Acidity

HCO₃^- <=> CO3(-2) + H+

Ka = 4.8 x 10-11

Ka = [CO3(-2)] x [H+] / [H₂CO₃ ]

4.8 x 10-11 =  [CO3(-2)] x 79 / (1.45 )

4.8 x 10-11 x 1.45 = [CO3(-2)] x 79

[CO3(-2)] = 8.8 x 10^-13 M = Alkalinity