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Find the normal [H2CO3] in blood H2CO3(aq) -><- H+(aq) +HCO3-(aq) (K=4.3x10^-7)

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Chemistry

Find the normal [H2CO3] in blood
H2CO3(aq) -><- H+(aq) +HCO3-(aq) (K=4.3x10^-7)

Answer & Explanation Solved by verified expert
4.3 Ratings (726 Votes)

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 , Ka = 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 Ka = ca2

==> a = √ ( Ka / c )

Given Ka = 4.3x10-7

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