Question

Sulfurous acid (H₂SO₃) has K_a1 = 1.500 × 10^-2 and K_a2 = 1.000 × 10^-7. Consider the titration of 60.0 mL of 1 M sulfurous acid by 1 M NaOH and answer the following questions.

What is the maximum number of protons that can sulfurous acid ionize (per molecule)?

b) Calculate the pH after the following total volumes of NaOH have been added. (Correct to 2 decimal places.)   No marks will be given if the number of decimal places is wrong.

i. 0.0 mL of NaOH  

ii. 30.0 mL of NaOH  

iii. 60.0 mL of NaOH  

iv. 90.0 mL of NaOH

v. 120.0 mL of NaOH

vi. 130.0 mL of NaOH

Answer 1

Sulfurous acid (H2SO3) has Ka1 = 1.500 × 10-2 and Ka2 = 1.000 × 10-7.

the titration of 60.0 mL of 1 M sulfurous acid by 1 M NaOH and answer the following questions.

What is the maximum number of protons that can sulfurous acid ionize (per molecule)?

it is a Diprotic acid, as it have acid dissociation constants, thus it may give 2 no. of protons per molecule), There are two -equivalence points in this titration.

b) Calculate the pH after the following total volumes of NaOH have been added. (Correct to 2 decimal places.)   No marks will be given if the number of decimal places is wrong.

i. 0.0 mL of NaOH  

we can neglect pKa2 initially ,

pH :  

[H+] = (1.500 × 10-2 * 1.0 ) ^0.5 = 0.1225 M

pH = -log [H+] = 0.91

ii. 30.0 mL of NaOH  

At this point solution behaves as buffer system :'

pH = pKa1 + log([HSO3-] /[H2SO3])

pKa1 = -log(ka1)

1M NaOH is used it is half ponit of 1st equivalence ponit , thus [HSO3-] = [H2SO3]

pH = pKa1 = 1.82

iii. 60.0 mL of NaOH  

it marks 1st equivalence point , at this point [HSO3-] is only species in solution,

it will have 2 equivilibriums ,

HSO3- (aq) <===> H+ (aq) + SO32- (aq)

HSO3- (aq)+ H2O (l) <===> OH- (aq) + H2SO3 (aq)

thus pH = 1/2 ( pKa1 + pKa2) = 1/2 ( 1.82 + 7.00) = 4.41

iv. 90.0 mL of NaOH :

At this point solution behaves as buffer system :'

pH = pKa2 + log( [SO32- ] / [HSO3-] )

pKa2 = -log(ka2)

1M NaOH is used it is half ponit of 1st equivalence ponit , thus [HSO3-] = [SO32- ]

pH = pKa2 = 7.00

v. 120.0 mL of NaOH :

2 equivalene point ,

at this point [SO32-] is only species in solution,

it will have equivilibriums ,

SO32- (aq)+ H2O (l) <===> OH- (aq) + HSO3- (aq)

pKb = 14 - pKa2 = 7

[SO32-] = 1M * 60 mL / ( 60 mL + 120 mL) = 0.333 M

[OH-] = (1.00 × 10-7 * 0.333 ) ^0.5 = 1.826*10-4 M

pH = 14 - pOH = 10.26

vi. 130.0 mL of NaOH

now NaOH decides pH ,

[OH-] = 10 ml* 1M / 190 mL = 0.05263 M

pH = 14 - pOH = 12.72