Question

For the diprotic weak acid H2A, Ka1 = 3.0 x 10-6 and K 2 = 9.0 x 10-9 What is the pH of a 0.0450 M solution of H,A? pH = What are the equilibrium concentrations of H, A and A2- in this solution? [H2A) = [A2-) =

Answer 1

The first dissociation of the weak acid H2A can be expressed as

H2A(aq) + Haq) + Hag

The equilibrium constant $K_{a1}$ for the first dissociation can be expressed as

HA](eq) x H Kai = (e) – 3.0 x 10-6 [H2A](eq)

Now, we can create the following ICE chart to calculate the equilibrium concentrations, given that starting concentration of the acid is 0.0450 M.

	$[H_2A]$	НА-	$[H^+]$
Initial, M	0.0450	0	0
Change, M	-x	+x	+x
Equilibrium, M	0.0450-x	x	x

From the ICE table, the Ka1 expression can be written as

-= 3.0 x 10-6 [HA-leg) [H+](eq) Kai = x X2 [H24] (eg) 0.0450 - = 2 = 3.0 x 10-6 x (0.0450 – 1) = 3,66 x 10-4 M

Hence, the equilibrium concentrations of different species after first dissociation of the acid are

H2A = 0.0450 M -I = 0.0450 M – 3.66 x 10-4 M -0.0446 M

HA ) = (+1= I = 3.66 x 10-4 M

Now, with the above equilibrium concentrations as starting concentrations, we can write the second dissociation of the acid as follows:

H Aſaq) → Haq) + Acryl

Hence, we can create the following equilibrium chart to calculate the equilibrium concentrations of different species after second dissociation

	НА-	$[A^{2-}]$	$[H^+]$
Initial, M	3.66 x 10-4	0	3.66 x 10-4
Change, M	-y	+y	+y
Equilibrium, M	3.66 x 10-4 - y	y	3.66 x 10-4 + 4

Hence, we can write the expression of Ka2 as follows:

[A' leg) [H+leg) Ka2 = [HA-(eg) →y 9.0 x 10-9 M yx (3.66 x 10-4 + y) 3.66 x 10-4 - y = 9.0 x 10-9

Note that

y<<3.66×10-4

Hence, the equilibrium concentrations of each species after second dissociation is

HAJ) = 3.66 x 10-4 M -y 3.66 x 10-4 M

H+1 = 3.66 x 10-4 M +y 3.66 x 10-4 M

42-1 = y = 9.00 x 10-9 M

Now, to answer the question, the pH of the solution can be calculated as follows:

pH = -log[H+1= -log(3.66 x 10-4) = 3.436 – 3.44

Hence, the pH of the solution is 3.44 approximately.

From above,

the equilibrium concentrations are

H2A = 0.0446 M

A-1 = 9.00 x 10-9 M

Note all answers are rounded up to 3 significant figures.