Question

Calculate the concentrations of all the species in a 0.301 M Na₂CO₃ solution.

[CO32−] =  M	[Na+] =  M
[HCO3−] =  M	[OH−] =  M
[H2CO3] =  × 10 M (Enter your answer in scientific notation.)	[H+] =  × 10 M (Enter your answer in scientific notation.)

Answer 1

Answer – We are given, [CO₃^2-] = 0.301 M

We know, Ka1 for the H₂CO₃ = 4.3*10^-7

Ka2 for the H₂CO₃ = 5.6*10^-11

Kb2 = 1.0*10^-14 /5.6*10^-11 = 1.79*10^-4

[Na₂CO₃] = 2 [Na⁺]

So, [Na⁺] = 2*0.301 M = 0.602 M

    CO₃^2- + H₂O-------> HCO₃^- + OH^-

I   0.301                          0              0

C    -x                              +x          +x

E 0.301-x                      +x           +x

Kb2 = [HCO₃^-][OH^-] / [CO₃^2- ]

1.79*10^-4 = x *x / 0.301-x

We can neglect x in the 0.301-x, since Kb value is small

So, 1.79*10^-4*0.301 = x²

So, x = 7.34*10^-3 M

So, x = [HCO₃^-] = [OH^-] = 7.34*10^-3 M.

[CO₃^2- ] = 0.301-7.34*10^-3 M

             = 0.294 M

Now second dissociation –

Kb2 = 1.0*10^-14 /4.3*10^-7= 2.33*10^-8

     HCO₃^- + H₂O-------> H₂CO₃ + OH^-

I   0.00734                       0              0

C    -x                                +x          +x

E 0.00734-x                   +x           +x

Kb2 = [H₂CO₃][OH^-] / [HCO₃^- ]

2.33*10^-8 = x *x / 0.00734-x

We can neglect x in the 0.00734-x, since Kb value is small

So, 2.33*10^-8*0.00734 = x²

So, x = 1.31*10^-5 M

So, x = [H₂CO₃] = 1.31*10^-5 M

[OH^-] = 1.31*10^-5 M

So, [HCO₃^- ] = 0.00734 - 1.31*10^-5 M

                       = 7.33*10^-3 M

Total [OH^-] = 7.34*10^-3 M + 1.31*10^-5 M

                   = 7.35*10^-3 M

So, [H⁺] = 1*10^-14 / 7.35*10^-3

               = 1.36*10^-12 M