Question

Calculate the pH of a 0.105 mol/L solution of carbonic acid. Answer is 3.668. How do I show my work for this question

Answer 1

Solution :-

Carbonic acid reaction with water

H2CO3 + H2O ------ > H3O+ + HCO3-

0.105                              0            0

-x                                   +x           +x

0.105-x                         x              x

ka= [H3O+][HCO3-]/[H2CO3]

4.3*10^-7 = [x][x]/[0.105-x]

since the ka is very small so we can neglect the x from denominator

4.3*10^-7 * 0.105 = x^2

4.515*10^-8 = x^2

taking square root of both sides

2.125*10^-4 = x= [H3O+]

formula to calculate the pH is as follows

pH= -log [H3O+]

pH= -log [2.125*10^-4]

pH= 3.67

the ka2 of the carbonic acid is not considered because it is very small so it is negligible.