Question

After performing Experiment 3, you have determined that your second equivalence point occurred after adding 42.50 mL of 0.1000 M HCl. Assuming that you dissolved your unknown sample in 150.00 mL of water, calculate the concentration of all species in solution AND the pH of the solution after 30.00 mL of HCl has been added? (Note: Kw = 1.00 x 10-14, Ka1= 4.44 x 10-7 and Ka2 = 4.69 x 10-11)

Answer 1

at second equivalence point only salt remians .

BH2+ concentration = 42.50 x 0.1 / (42.50 + 150) = 0.0221 M

BH2+ ------------------------> BH+ + H+

0.0221 - x x x

Ka2 = x^2 / 0.0221 -x

4.69 x 10-11 = x^2 / 0.0221 -x

4.69 x 10-11 = x^2 / 0.0221 -x

x = 1.018 x 10^-6

[BH2+] = 0.0221- x

[BH2+] = 0.02209 M

[BH+] = 1.018 x 10^-6 M

[H+] = 1.018 x 10^-6

[OH-] = Kw / [H+] = 1.0 x 10^-14 / 1.018 x 10^-6

[OH-] = 9.82 x 10^-9 M

B + H2O --------------------> BH+ + OH-

Kb1 = [BH+][OH-]/[B]

Kw / Ka1 =  [BH+][OH-]/[B]

(1.0 x 10^-14 ) / (4.44 x 10^-7) =  [BH+][OH-]/[B]

2.25 x 10^-8 =  [BH+][OH-]/[B]

[B] = [BH+][OH-] /2.25 x 10^-8

[B] = 4.44 x 10^-7 M