Question

25.00 ml of 0.100 M H2A is being titrated with 0.100 M NaOH. (Ka1=1.3x10^-2 Ka2=5.9x10^-7) Calculate the pH at 0ml NaOH, 5ml NaOH, 25ml NaOH, 25.5 ml NaOH, 50ml NaOH, and 51ml NaOH.

Answer 1

When 0ml of NaOH is added

H2A---> HA- + H+ Ka1= 1.3X10-2 and HA----- > H+ + A- ka2= 5.9*10-7

Ka1= [ HA-] [H+]/ [H2A]

let x moels be dissociated

At equilibrium

H2A =0.1-x

HA- = [H+] =x

x2/(0.1-x)= 1.3*10-2

x2 =1.3*10^-2*0.1- 1.3*10^-2x

x²=1.3*10-3 -1.3*10-2x

x2+1.3*10^-2x-1.3*10^-3= 0

x =0.03017

[H+] =0.03017

For the second case

HA- ---> H+ + A-

Ka2= [H+] [A-]/ [HA-]

Let x be the amount dissociated through second reaction

5.9*10^-7= [H+] [A-]/ (0.03017-x)

5.9*10-7= x²/(0.03017-x)

x²= 5.9*10^-7 (0.03017-x)

x2= 0.177807*10-7- 5.9*10-7x

x2+ 5.9*10^-7x-0.177807*10^-7=0 x =0.000133305

[H+] =0.000133305

pH= -log [H+] =3.87

When 5ml of NaOH is added

the reaction between NaOH and H2A can be written as

H₂A+ 2NaOH---Na₂A+ 2H₂O

Moels of H2A= 0.1* 25/1000=0.0025 moles

Moels of 0.1vNaOH= 5*0.1/1000=0.0005

Moles of NaOH gets consumed completely

and moles of H2A consumed as per the reaction are 0.0005/2 =0.00025

moels of H2A unreacted = 0.0025- 0.00025=0.00225

volume of the mixture =25+5 =30ml concentration =0.00225*1000/30=0.075 M

for the first reaction H2A---> HA- + H+

let x be the dissociation

x²/(0.075-x) = 1.3*10-2

x2=0.000975- 1.3*`10^-2 x , x2 +1.3*10-2x -0.000975= 0, x =0.025494

for the second reaction HA- ----> H+ + A-

x2/(0.025494-x) = 5.9*10-7

x²= 1.50*10^-8- 5.9*10-7x

x²+5.9*10^-7x-1.5*10^-8 =0 x =0.000122349 pH= 3.91

3. Moels of NaOH =(25/1000)*0.1 = 0.0025

Moels of H2A =0.0025

moels of H2A that can neutralize 0.0025 moles of NaOH =0.0025/2 =0.00125

remaining moles - 0.0025-0.00125= 0.00125

volume of mixture= 25+25 =50 ml concentration of H2A =0.00125*1000/50 =0.025 moles

x2./(0.025-x)= 1.3*10-2

x2=0.025*1.3*10-2 -1.3*10-2x

x2= 0.000325-1.3*10-2x

x2+1.3*10-2x -0.000325=0 x =0.012664

for the second reaction HA- > H+ + A-

x²/(0.012664-x)= 5.9*10-7

x2 =5.9*10^-7 *0.012664- 5.9*10-7x

x2+5.9*10^-7x -7.471*10^-9=0 x= 8.58*10-5

pH=-log [H+] =4.066

4. Moles of NaOH in 25.5 ml = 0.1*25.5/1000= 0.00255

Moels of H2A to neutralize the NaOH =0.00255/2 =0.001275

Balance H2A =0.025-0.001275 = 0.001225 and its concentration =0.001255*1000/50.5 =0.02425

x2/(0.02425-x)= 1.3*10-2

x2= 1.3*10^-2*0.02425- 1.3*10-2x

x2+1.3*10-2x -0.00031525 x= 0.0124

for the second case

x2/(0.0124-x) =5.9*10-7

x2 =5.9*10-7*0.0124-5.9*10-7x

x2+5.9*10^-7x-7.32*10^-9= 0 x= 8.49*10-5 pH= -log(x)= 4.07

when 50ml NaOH is added moles of NaOH =50*0.1/1000 =0.005

Moels of H2A consumed =0.005/2 = 0.0025 all the H2A reacts and hence pH= 7