What is the Ka of a 0.0675 M solution of carbonic acid
(H2CO3) with a pH of 5.19?
Ka = [ ? ] x 10
Help ASAP

The Ka is 6.183 * 10^-10

Explanation:

The first thing to do here is to write the dissolution equation of carbonic acid in water.

H2CO3 + H20 ———> H30+ + HCO3-

After writing this, the next thing is to write the expression for Ka

Mathematically that would be ;

Ka = [H30+][HCO3-]/[H2CO3]

Next thing is to set up an ICE table

ICE table stands for initial, change and equilibrium.

For clarity sake, please check attachment for the ICE table for this chemical reaction

Mathematically by definition;

pH = -log [H3O+]

From the question, we were made to know that the pH is 5.19

Hence;

5.19 = -log [H3O+]

-5.19 = log [H30+]

[H3O+] = 10^-(5.19)

[H3O+] = 6.46 * 10^-6 M

So how does this relate to the ICE table?

From the ICE table, we can see that the equilibrium concentration of the hydroxonium ion equals x.

This means that the equilibrium concentration of the hydroxonium ion [H3O+] , demoted as x = 6.46 * 10^-6 M

Now according to the Ka equation;

Kindly recall that;

Ka = [H30+][HCO3-]/[H2CO3]

Now, using their equilibrium concentrations;

Ka = (x)(x)/(0.0675-x)

Ka= x^2/(0.0675-x)

Kindly substitute 6.46 * 10^-6 for x

Ka = (6.46 * 10^-6)^2/(0.0675-(6.46 * 10^-6)

Ka = 6.183 * 10^-10

the answer is b & c i believe : )