Question

Calculate the concentration of all species in a 0.175 M solution of H2CO3.

Enter your answers numerically separated by commas. Express your answer using two significant figures.

[H2CO3], [HCO3^-], [CO3^2-], [H3O^+], [OH^−] =

Answer 1

                     H2CO3 + H2O àà   H3O⁺     +    HCO₃^-

Initial                   0.175------------------[0]----+-----[0]

Equilibrium            [-x]-------------------[x]-----+-----[x]

Final                 [0.175-x]-----------------[x]-----+-----[x]

Now K_a1 = [H⁺ ][ HCO3^-]/[H2CO3]

We know that K_a1 = 4.4 × 10^-7 and K_a2 = 4.7 × 10^-11 for H2CO3 acid,

By substituting values we get ,

4.4 × 10^-7 = [x][x]/ [0.175-x]

Neglecting denominator we get x= 6.6332 × 10^-4

Therefore concentration of [H3O⁺] and [HCO3^-] = 6.6332 × 10^-4 M-----------------1

Final Concentration of [H2CO3] can be calculated as

[H2CO3] = [0.175-x] = 0.175 - 6.6332 × 10^-4 = -0.174 M---------------------------------2

HCO₃^- further dissociates as follows,

  HCO₃^-    +   H2O   àà   H3O⁺ + CO₃^2-

Ka2 = [H3O⁺ ][ CO₃^2-]/[ HCO₃^-]

Form equation 1 we can write that (H3O⁺) = (HCO3^-)

4.7 × 10^-11 = 6.6332 × 10^-4 × [ CO₃^2-]/6.6332 × 10^-4

[ CO₃^2-] = 4.7 × 10^-11 M

Concentration of [ CO₃^2-] = 4.7 × 10^-11 M------------------------------------------------------3

H3O+ dissociates as follows,

pH = -log [H+]

pH= -log 6.6332 × 10^-4 = -(-3.1782) = 3.178

Now, pH+pOH = 14

pOH= 14-3.178= 10.822

pOH = -Log [OH^-]

[OH^-] = antilog 10.822

Concentration of [OH^-] = 1.506 ×10^-11 M--------------------------------------------4