For the diprotic weak acid H2A, a1=3.2×10−6 and a2=6.1×10−9 .

What is the pH of a 0.0750 M solution of H2A ?
What are the equilibrium concentrations of H2A and A2− in this solution?

Answers

Answer 1

In the first dissociation of H2A:

molarity    H2A(aq)↔ (HA)^-(aq) + H^+(aq)

initial                0.05 m          0 m           0 m

change               -x                 +x               +x

equilibrium    0.05-x               x                 x

we can neglect X in [H2A] as it so small compared to the 0.05

so by substitution in Ka equation:

Ka1 = [HA][H] / [H2A]

2.2x10^-6 = X^2/0.05

X = √(2.2x10^-6)*(0.05)= 1.1x10^-7

X=   3.32x10^-4 m

∴ [H2A] = 0.05 - 3.32x10^-4 = 0.0497 m

[HA] = 3.32x10^-4 m

[H] = 3.32x10^-4 m

the second dissociation of H2A: when ka2 = 8.2x10^-9

                          HA-(aq)     ↔ A^2- (aq) + H+(aq)

at equilibrium   3.32x10^-4        y              3.32x10^-4

Ka2           = [H+][A^2-] / [HA]

8.2x10^-9 = Y(3.32x10^-4)/(3.32x10^-4)

∴y = 8.2x10^-9 m

∴[A] = 8.2x10^-9 m

PH= -㏒[H+]

   = -㏒(3.32x10^-4)= 3.479  

[A]=8.2x10^-9 m

[H2A] = 0.0497 ≈ 0.05 m


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Problem solution

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my sheesh hurts my sheesh hurtsmy sheesh hurtsmy sheesh hurtsmy sheesh hurtsmy sheesh hurtsmy sheesh hurtsmy sheesh hurtsmy sheesh hurtsmy sheesh hurtsmy sheesh hurtsmy sheesh hurtsmy sheesh hurtsmy sheesh hurtsmy sheesh hurtsmy sheesh hurts

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