Question

Calculate the concentration of all species in a 0.155 M solution of H2CO3.

Express your answer using two significant figures.

[H2CO3], [HCO−3], [CO2−3], [H3O+], [OH−] =		M

Answer 1

Answer - Given, [H₂CO₃] = 0.155 M and Ka1 = 4.3*10^-7 , Ka2 = 5.6*10^-11

We know H₂CO₃ is weak acid, so we need to put ICE chart for the first dissociation -

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

I    0.155                             0                0

C     -x                                +x            +x

E    0.155-x                       +x             +x

Ka1 = [HCO₃^- ] [H₃O⁺] / [H₂CO₃]

4.3*10^-7 = (x) * (x) / (0.155-x)

We can neglect x in 0.155-x , since Ka1 value is too low

4.3*10^-7 * 0.155 = x²

x = 0.000258 M

[H₃O⁺] = x = 0.000258 M

x = [HCO₃^-] = 0.000258 M

[H₂CO₃] = 0.155-x

               = 0.155 – 0.000258

               = 0.1547 M

Now second dissociation-

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

I    0.000258                      0            0

C      -x                            +x          +x

E 0.000258–x                  +x           +x

Ka2 = [CO₃^2-] [H₃O⁺] / [HCO₃^-]

5.6*10^-11 = (x) * (x) / (0.000258-x)

We can neglect x in 0.000258-x , since Ka2 value is too low

5.6*10^-11 * 0.000258 = x²

x = 1.20*10^-7 M

[HCO₃^-] = 0.000258 -x

           = 0.000258- 1.20*10^-7 M

         = 2.58*10^-4 M

x = [H₃O⁺] = 1.20*10^-7 M

x = [CO₃^2-] = 1.20*10^-7 M

total [H₃O⁺] = 0.000258 M + 1.20*10^-7 M

                  = 2.58*10^-4 M

We know,

[H₃O⁺] [OH^-] = 1.0*10^-14

So, [OH^-] = 1.0*10^-14 / 2.58*10^-4 M

                = 3.87*10^-11 M