Question

H2SO4 is a strong acid because the first proton ionizes 100%. The Ka of the second proton is 1.1x10-2. What would be the pH of a solution that is 0.100 M H2SO4? Account for the ionization of both protons.

Answer 1

Question options:

a) 2.05

b) 0.963

c) 0.955

d) 1.00

Answer:

b) 0.963

Step-by-step explanation:

H2SO4→ HSO4- + H3O+

HSO4- + H2O ⇌ SO42- + H3O+

Construct ICE table:

HSO4- (aq) + H2O ⇌ SO42- (aq) + H3O+ (aq)

I 0.1 solid & 0 0.1

C -x liquid + x + x

E 0.1 - x are ignored x 0.1 + x

Calculate x

Ka = products/reactants

=
`([SO42-] [H3O+])/([HSO4-])`

0.011 =
`(x (0.1 + x))/(0.1 - x)`

0.011 x (0.1 -x) = o.1x + x^2

0.0011 - 0.011 x - o.1x - x^2 = 0

0.0011 - 0.011 x - x^2 = 0

Use formula to solve for quadratic equation

x =
`{ -b +,-\sqrt{b^2 - 4ac` / 2a

a = -1, b = -0.111, c = 0.001

Solve for x

x =
`\sqrt[-(-o.111)]{(-0.111)^2 - 4(-1) (0.0011) }` / 2(-1)

x = 0.111 +,-
`√(0.012321 + 0.0044)` / -2

x = 0.111 +,-
`√(0.016721)` / -2

x =
`(0.111 +, - 0.1293)/(-2)`

x =
`(0.111 + 0.1293)/(-2)` , x =
`(0.111 - 0.1293)/(-2)`

x =
`(0.2403)/(-2)` , x =
`(0.0183)/(-2)`

x = - 0.12015 , x = 0.00915

x cannot be negative, so

x = 0.00915 M

Calculate [H3O+]

[H3O+] = 0.1 M + x

[H3O+] = 0.1 M + 0.00915 M

[H3O+] = 0.10915 M

Clculate pH

pH = - log [ H3O+]

pH = - log [ 0.10915]

pH = 0.963