Question

The pH of the water is measured as 8.5 and the total concentration of carbonate species is found to be 0.00180M. Find the concentrations of H2CO3, HCO3-and CO32- by both graphically and using the equations

Answer 1

Sicne the total carbonate concentration is 0.00180 M, the initial concentration of H2CO3 was 0.00180 M.

If 'x' and 'y' are respectively the first and second degree of dissociation of H2CO3, then

For the first dissociation of H2CO3

---------------------H2CO3 <---------> HCO₃^-(aq) + H₃O⁺ ; Ka1 = 4.3*10^-7

Init.Conc(M): 0.00180 -------------- 0 --------------- 0

Eqm.conc(M):0.00180(1 - x),   0.00180x, 0.00180x

Ka1 = 4.3*10^-7 = (0.00180x)*(0.00180x) / 0.00180(1 - x) = 0.00180*x² / (1 - x)

=> x = 0.0155

Hence [H2CO3] = 0.00180(1 - x) = 0.00180(1 - 0.0155) = 0.00177 M (answer)

[HCO₃^-(aq)] after 1st dissociation = 0.00180x = 0.00180*0.0155 = 2.79*10^-5 M

For the second dissociation of H2CO3

---------------------HCO₃^-(aq) <---------> CO₃^2-(aq) + H₃O⁺ ; Ka2 = 4.8*10^-11

Init.Conc(M): 2.79*10^-5 ----------------- 0 --------------- 2.79*10^-5 M

Eqm.conc(M):2.79*10^-5(1 - y), 2.79*10^-5y, (2.79*10^-5y +2.79*10^-5)

Ka2 = 4.8*10^-11 = (2.79*10^-5y +2.79*10^-5)*2.79*10^-5y / 2.79*10^-5(1 - y) 2.79*10^-5*y

=> y =  1.72*10^-6  

Hence [HCO₃^-(aq)] = 2.79*10^-5(1 - y)   2.79*10^-5 (answer)

[CO₃^2-(aq)] =  2.79*10^-5y =  2.79*10^-5* 1.72*10^-6 = 4.8*10^-11 (answer)