Question

Calculate the pH at 0, 25.0, 50.0, 75.0, 100, and 140% titration in the titration of both protons of the diprotic acid H2SO3 with 0.100M NaOH, starting with 25.0 mL of 1.00 M H2A.

Ka1=1.20*10^-2

Ka2=6.60*10^-8

Answer 1

I think it is H2SO4

Following reation occr

H2SO4----->H⁺+HSO4^-

HSO4^-------->H⁺+SO4^2-

1.

NaOH=0

Ka1=[H⁺][HSO₄^-]/[H₂SO4]=x*x/1=1.20*10^-2

x=0.1095

[H⁺]=0.1095

pH=-log[H⁺]=0.960

2. NaOH=25%

1st buffer region, both H2A and OH- present

Ka1 > >Ka2, second equilibrium makes little contribution, pH calculated like a normal buffer solution

So half way to equivalence

[H2SO4]=[OH^-]

So

[H⁺]=Ka1=1.20*10^-2

pH=1.92

3.

50%

solution is like that of a salt of a diprotic acid

[H⁺]=((Ka₂*[NaOH]+Kw)/(1+[NaOH]/Ka1))^0.5

Kw=10^-14

[H⁺]=2.66*10^-5

pH=4.575

(4)

75%

pH in 2nd buffer region

2nd buffer region, both HA- and A2- present if Ka1 > 103 Ka2, first equilibrium makes little contribution, pH calculated like a normal buffer solution half way to equivalence,[HSO3^-]=[SO4^2-]

[H⁺]=Ka2=6.60*10^-8

pH=7.18

(5)

100%

pH at the second equivalence point Like a salt of A^2-, main equilibrium is

SO₄^2-+H2O---->OH^-+HSO4^-

K_b1=Kw/Ka2=10^-14/6.60*10^-8=1.515*10^-7

[OH^-]=(Kb1*1)^0.5=3.892*10^-4

pOH=3.40977

pH=70.59

(6)

140%

pH beyond 2nd equiv. pt.

treated like the addition of strong base to water.

pH=pw=14