Question

Find the normal [H2CO3] in blood

H2CO3(aq) -><- H+(aq) +HCO3-(aq) (K=4.3x10^-7)

Answer 1

Let a be the dissociation of the weak acid,H₂CO₃
                          H₂CO₃ <---> H ⁺ + HCO₃^-

initial conc.            c               0        0

change                -ca            +ca   +ca

Equb. conc.         c(1-a)          ca      ca

Dissociation constant , K_a = ca x ca / ( c(1-a)

                                         = c a² / (1-a)

In the case of weak acids a is very small so 1-a is taken as 1

So K_a = ca²

==> a = √ ( K_a / c )

Given K_a = 4.3x10^-7

If we know the dissociation of H₂CO₃ (a) then we calculate the concentration of H₂CO₃