Question

Consider a 0.0100 M solution of malonic acid (H2A) with pKa1 = 2.847 and pKa2 = 5.696

a. Calculate the pH of the solution.

b. Calculate the fraction of malonic acid existing as A^2-. Does the second dissociation step make a significant contribution to the H+ in solution?

Answer 1

Ans :-

a). pK_a1 = 2.847

K_a1 = 10^-pK_a1 = 10^-2.847 = 1.42 x 10^-3

and

pK_a2 = 2.847

K_a2 = 10^-pK_a2 = 10^-5.696 = 2.01 x 10^-6

ICE table is :

............................H₂A (aq) <-----------------------------> H⁺ (aq) .........................+............................HA^- (aq)

Initial...................0.0100 M.........................................0.0 M............................................................0.0 M

Change.................-y.....................................................+y.................................................................+y

Equilibrium..........(0.0100-y) M.......................................y M...............................................................y M

Expression of acid dissociation constant i.e. K_a(which is equal to the product of the molar concentration of products divided by product of the molar concentration of reactants raise to power their stoichiometric coefficient when reaction is at equilibrium stage).

K_a1 = [H⁺].[HA^-]/[H₂A]

1.42 x 10^-3 = y²/(0.0100-y)

y² + 1.42 x 10^-3 y - 1.42 x 10^-5 = 0

On solving

y = 0.00312

Therefore,

Equilibrium concentration of H⁺= y = 0.00312 M

Equilibrium concentration of HA^-  = y = 0.00312 M

Because, [H⁺] = 0.00312 M

pH = - log[H⁺]

= - log 0.00312 M

= 2.51

Therefore, pH = 2.51

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b). ICE table is :

............................HA^-(aq) <-----------------------------> H⁺ (aq) .........................+................................A^2- (aq)

Initial...................0.00312 M.........................................0.0 M............................................................0.0 M

Change.................-y.....................................................+y.................................................................+y

Equilibrium..........(0.00312-y) M.......................................y M...............................................................y M

Expression of acid dissociation constant i.e. K_a(which is equal to the product of the molar concentration of products divided by product of the molar concentration of reactants raise to power their stoichiometric coefficient when reaction is at equilibrium stage).

K_a2 = [H⁺].[HA^-]/[H₂A]

2.01 x 10^-6  = y²/(0.00312-y)

y² + 2.01 x 10^-6  y - 6.27 x 10^-9 = 0

On solving

y = 8.0 x 10^-5

Therefore,

Fraction of malonic acid = y = 8.0 x 10^-5 M

Equilibrium concentration of H⁺= y = 8.0 x 10^-5 M

No, because, the concentration of H⁺ in the second dissociation step is very small therefore, second dissociation step does not make a significant contribution to the H+ in solution.