Question

What is the pH of a solution that is 0.0924 M in malonic acid? What will be the equilibrium concentration of each of the three malonic acid species? pKa1 = 2.85 pKa2 = 5.70

pH: _________________
[H2A]: ____________________ [HA-]: _________________ [A2-]: _________________

Answer 1

Ans :-

Given molar concentration of H₂A = 0.0924 M

pK_a1 = 2.85

So,

K_a1 = 10^-pK_a1 = 10^-2.85 = 1.41 x 10^-3

Similarly,

pK_a2 = 5.70

So,

K_a2 = 10^-pK_a2 = 10^-5.70 = 2.00 x 10^-6

Now, ICE table of H₂A is :

..................................H₂A <-----------------------------------> H⁺.............................+.......................HA^-

Initial .........................0.0924 M.........................................0 M......................................................0 M

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

Equilibrium.............(0.0924-y) M........................................y M......................................................y M

Where, y = Amount dissociated per mole

Expression of equilibrium constant i.e. K_a1(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]

1.41 x 10^-3 = y² / (0.0924-y)

y² + 1.41 x 10^-3y - 1.30284 x 10^-4 = 0

On solving

y = 0.0107

So,

[H⁺] = y = 0.0107 M

[HA^-] = y = 0.0107 M

and

[H₂A] = 0.0924 - 0.0107 = 0.0817 M

again,

ICE table of HA^- is :

..................................HA^- <-----------------------------------> H⁺.............................+.......................A^2-

Initial .........................0.0107 M.........................................0.0107 M..........................................0 M

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

Equilibrium.............(0.0107-y) M........................................(0.0107+y) M..................................y M

Where, y = Amount dissociated per mole

Expression of equilibrium constant i.e. K_a2(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⁺].[A^2-] / [HA^-]

2.00 x 10^-6 = y(0.0107+y) / (0.0107-y)

As, y <<<0.0107, o neglecy y as compare to 0.0107

Therefore,

y = 2.00 x 10^-6

So,

[H⁺] = 0.0107 + 2.00 x 10^-6 = 0.0107 M

As,

pH = - log [H⁺]

pH = - log 0.0107 M

pH = 1.97

Therefore, pH = 1.97

[H2A] =  0.0817 M

[HA^-] = 0.0107- 2.00 x 10^-6 = 0.0107 M

So,

[HA-] = 0.0107 M

and

[A2-] = 2.00 x 10-6 M