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

Find the normal [H2CO3] in blood

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

Expert Solution

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

queen_honey_blossom answered 1 year ago

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