Question

What is the pH of a 0.65 M solution of carbonic acid?

Report your answer to 2 decimal places.

Answer 1

Carbonic acid – H₂CO₃ . It is a weak diprotic acid.

H₂CO₃ = H⁺ + HCO₃^- , Ka1 = 2.5 x 10^-4

HCO₃^­- = H⁺ + CO₃^2- , Ka2 = 5.6 x 10^-11

A 0.65 M H₂CO₃ solution is taken. Let x M of H₂CO₃ dissociate to form x M H⁺ + x M HCO₃^- ( according to the dissociation for Ka1)

The following ICE table can be constructed :

Concentration ( M )	H2CO3	H+	HCO3-
Initial	0.65	0	0
Change	- x	+ x	+ x
Equilibrium	0.65 - x	x	x

Now, this x M of HCO₃^- formed can further dissociate according to the equation for Ka2. Let y M of HCO₃^_ dissociate. The ICE table is:

Concentration ( M )	HCO3-	H+	CO32-
Initial	x	x	0
Change	- y	+ y	+ y
Equilibrium	x - y	x + y	y

Now, the expressions for Ka1 and Ka2 are:

Ka1 = = 2.5 x 10^-4 .

Or, = 2.5 x 10^-4 .

And, Ka2 = = 5.6 x 10^-11 .

Or, = 5.6 x 10^-11 .

Approximation made : Since Ka2 <<<< Ka1, it is only the first dissociation that is relevant. So, y is negligible.

So, modifying the equation of Ka1 and Ka2 by assuming : x >>> y, So, ( x + y ) = x and ( x-y) = x.

= 2.5 x 10^-4 . Solving for x , x = 0.01262 [**detailed solving at the end]

And = 5.6 x 10^-11 . Solving for y, y = 5.6 x 10^-11

As initially assumed, x >>> y.

Now, equilibrium concentration of H+ = x + y = 0.01262 + 5.6 x 10^-11 . So, [H+] = 0.01262 M

pH - -log[H+] = -log(0.01262 ) = 1.90

---------**solving-------------

= 2.5 x 10^-4

Or, x² = (2.5 x 10^-4 ) ( 0.65 – x)

Or, x² = 1.625 x 10^-4 - 2.5 x 10^-4 x

Or, x² + 2.5 x 10^-4 x – 1.625 x 10^-4 = 0. Quadratic equation.

So, ,

Or,

Taking positive solution, x = 0.01262