In the titration of 30.00 mL of 0.100 M sulfurous acid (H_2 SO_3) by 0.100 M sod
ID: 919174 • Letter: I
Question
In the titration of 30.00 mL of 0.100 M sulfurous acid (H_2 SO_3) by 0.100 M sodium hydroxide, you should almost be able to immediately identify the volume and pH for three points along the titration curve (no ICE table required - only show work for these three points if you need to). Identify these three and then determine the pH at the start, Vb = 0 (no base added so just the acid in solution, ice table recommended) and the pH 10.00 mL after the second endpoint. H_2 SO_3 (aq) + H_2 O doubleheadarrow H_3 O^+ (aq) + HSO_3^+ (aq) HSO_3^+ (aq) + H_2 O doubleheadarrow H_3 O^+ (aq) + SO_3^2- (aq)Explanation / Answer
moles of H2SO3 = 0.1 M x 0.03 L = 3 x 10^-3 mols
(a) First point,
No base added Vb = 0
H2SO3 <==> H+ + HSO3-
let x amount has dissociated,
Ka1 = 5.01 x 10^-4 = x2/0.1
x = [H+] = 7.08 x 10^-3 M
pH = -log[H+] = 2.15
(b) Second point,
Vb = 3 x 10^-3 mols/0.1 = 0.03 L = 30 ml of NaOH added
HSO3- hydrolyzes as,
HSO3- + H2O <==> H2SO3 + OH-
molarity of HSO3- = 0.1 M x 0.03 L/0.06 L = 0.05 M
let x amount has hydrolyzed then,
Kb1 = Kw/Ka1 = [H2SO3][OH-]/[HSO3-]
1 x 10^-14/5.01 x 10^-4 = x^2/0.05
x = [OH-] = 9.99 x 10^-7 M
pOH = 6.0
pH = 14 - pOH = 8.0
(c) Third point - second end point
Vb = 2 x 3 x 10^-3 mols/0.1 = 0.06 L = 60 ml of NaOH added
SO3^2- hydrolyzes as,
SO3^2- + H2O <==> HSO3- + OH-
molarity of SO3^2- = 0.1 M x 0.03 L/0.09 L = 0.033 M
let x amount has hydrolyzed then,
Kb1 = Kw/Ka1 = [H2SO3][OH-]/[HSO3-]
1 x 10^-14/6.607 x 10^-8 = x^2/0.033
x = [OH-] = 7.07 x 10^-5 M
pOH = 4.15
pH = 14 - pOH = 9.85
Now,
(d) pH after 10 ml of 0.1 M NaOH added
moles of NaOH = 0.1 x 0.01 = 1 x 10^-3 mols
remaining moles of H2SO3 = 2 x 10^-3 mols
[HSO3-] = 1 x 10^-3/0.04 = 0.025 M
[H2SO3] = 2 x 10^-3/0.04 = 0.05 M
pH = pKa1 + log([base]/[acid])
= 3.30 + log(0.025/0.05)
= 3.0