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

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

  = $\frac{[SO42-] [H3O+]}{[HSO4-]}$

0.011 = $\frac{x (0.1 + x)}{0.1 - x}$

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 +,- $\sqrt{0.012321 + 0.0044}$ / -2

x = 0.111 +,- $\sqrt{0.016721}$ / -2

x = $\frac{0.111 +, - 0.1293}{-2}$

x = $\frac{0.111 + 0.1293}{-2}$   , x = $\frac{0.111 - 0.1293}{-2}$

x = $\frac{0.2403}{-2}$    , x = $\frac{0.0183}{-2}$

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