Question

 

1. Calculate the pH and the concentrations of all species present in a .0230 M HPO₄^- (K_a = 4.90 x 10^-10)

2. Calculate the pH and the concentrations of all species present in a 3.20 x 10^-5 M CH₃CO₂H (K_a = 1.70 x 10^-5)

Answer 1

1.

HPO₄^-2    H⁺ + PO₄^3-

I: 0.0230 0 0
C: -x +x +x
E: 0.0230-x x x

Ka =  [H⁺] [PO₄^3-] / [HPO₄^-2]

4.90 x 10^-10 = (x) (x) / (0.0230-x)

4.90 x 10^-10 = (x) (x) / 0.0230 (since Ka is very small, the x term in denominator can be dropped)

x² = 1.13 x 10^-11

x = 3.36 x 10^-6

So, [H⁺] = x = 3.36 x 10^-6

Now,

[H⁺] = [PO₄^3-] = 3.36 x 10^-6 M

[HPO₄^-2] = 0.0230 - x = 0.0230 - 0.00000336 = 0.02299664 M

pH = - log [H⁺]

= - log (3.36 x 10^-6)

= 5.47

2.

CH₃COOH    H⁺ + CH₃COO^-

I: 0.000032 0 0
C: -x +x +x
E: 0.000032-x x x

Ka =  [H⁺] [CH₃COO^-] / [CH₃COOH]

1.70 x 10^-5 = (x) (x) / (0.000032-x)

1.70 x 10^-5 = (x) (x) / 0.000032 (since Ka is very small, the x term in denominator can be dropped)

x² = 5.44 x 10^-10

x = 2.33 x 10^-5

So,  [H⁺] = x = 2.33 x 10^-5

pH = - log [H⁺]

= - log (2.33 x 10^-5)

= 4.63

Now,

[H⁺] = [CH₃COO^-] = 2.33 x 10^-5 M

[CH3COOH] = 3.20 x 10^-5 M - x = 3.20 x 10^-5 - 2.33 x 10^-5 M = 8.70 x 10^-6 M