Question

For each of the following, is the solution acidic, basic, neutral, or cannot be determined? For each, write the equation for the dominant equilibrium which determined the pH, and justify your pH prediction. Kw = 1.0 x 10-14

100 mL of 0.10 M NaH3P2O7; Ka1 = 3.0 x 10-2, Ka2 = 4.4 x 10-3, Ka3 = 2.5 x 10-7, and Ka4 = 5.6 x 10-10 for H4P2O7.

100 mL of 0.10 M K2H2P2O7; see Part a for Ka values

100 mL of 0.10 M Li3HP2O7; see Part a for Ka values

100 mL of 0.10 M Na4P2O7; see Part a for Ka values

100 mL of 0.10 M C9H7N; Kb = 6.3 x 10-10 for C9H7N

100 mL of 0.10 M Ba(ClO4)2

100 mL of 0.10 M CH3CH2OH

Answer 1

1) H3P2O7-(aq) + H2O(l) <-------> H4P2O7(aq) + OH-(aq)

Kb = KW/Ka = 1×10^-14/3×10^-2 = 3.3×10^-13

3.3 ×10^-13 =[H4P2O7][OH]/[H3P2O7-]

[OH-] = 1.82×10^-7

pOH = 6.73

pH = 14 - 6.73 = 7.27

So ,the solution is near to neutral

2) H2P2O72- + H2O < -------> H3P2O7- + OH-

Kb = 1×10^-14/4.4 ×10^-3 =2.3 ×10^-12

2.3×10^-12 = [H3P2O7-][OH-]/[H2P2O7]

[OH-] = 4.8 × 10^-7

pOH = 6.32

pH = 7.68

The solution is slightly basic

3) HP2O7^3- + H2O ---------> H2P2O7^2- + OH-

Kb = 1× 10^-14/2.5 ×10^-7 = 4×10^-8

4×10^-8 = [H2P2O7][OH-]/[HP2O7^3-]

[OH-] = 6.3 × 10^-5

pOH = 4.2

pH = 9.8

The solution is basic

4) P2O7^4- + H2O ------> HP2O7^3- + OH-

Kb = 1×10^-14/5.6 ×10^-10 = 1.8×10^-5

1.8 × 10^-5 = [HP2O7^3-][OH-]/[P2O7^4-]

[OH-] = 1.3 × 10^-3

pOH = 2.88

pH = 11.12

The solution is basic