What are the concentrations of H_2CO_3H2CO3 and its conjugate bases in a raindrop with pH = 5.90pH=5.90?

Calculate the concentrations of carbonic acid, bicarbonate ion (HCO_3^-HCO−3) and carbonate ion (CO_3^(2−)CO2−3) that are in a raindrop that has a pH of 5.90, assuming that the sum of all three species in the raindrop is 1.0*10^(−5)M1.0⋅10−5M.

Where,

K_(a_1) = 4.3*10^-7Ka1=4.3⋅10−7, and
K_(a_2) = 5.6*10^-11Ka2=5.6⋅10−11

1 Answer
Al E.
Dec 8, 2017

We must assume,

[H_2CO_3] + [HCO_3^(-)] + [CO_3^(2-)] = 1.0*10^-5[H2CO3]+[HCO−3]+[CO2−3]=1.0⋅10−5, and
pH = 10^(-5.90) approx 1.26*10^-6MpH=10−5.90≈1.26⋅10−6M

Moreover,

H_2CO_3(aq) rightleftharpoons H^(+)(aq) + HCO_3^(-)(aq)H2CO3(aq)⇌H+(aq)+HCO−3(aq)
HCO_3^(-)(aq) rightleftharpoons H^(+)(aq) + CO_3^(2-)(aq)HCO−3(aq)⇌H+(aq)+CO2−3(aq)

and,

K_(a_1) = ([H^(+)][HCO_3^(-)])/([H_2CO_3])Ka1=[H+][HCO−3][H2CO3]
K_(a_2) = ([H^(+)][CO_3^(2-)])/([HCO_3^(-)])Ka2=[H+][CO2−3][HCO−3]

So, we will relate these equilibrium expressions,

[H_2CO_3] = ([H^(+)][HCO_3^(-)])/(K_(a_1)[H2CO3]=[H+][HCO−3]Ka1

[CO_3^(2-)] = (K_(a_2)[HCO_3^(-)])/([H^(+)])[CO2−3]=Ka2[HCO−3][H+]

and our assumption,

([H^(+)][HCO_3^(-)])/(K_(a_1)) + [HCO_3^(-)] + (K_(a_2)[HCO_3^(-)])/([H^(+)]) = 1.0*10^-5[H+][HCO−3]Ka1+[HCO−3]+Ka2[HCO−3][H+]=1.0⋅10−5

Now, to simplify a bit,

[H^+]^2[HCO_3^(-)] + K_(a_1)[H^+][HCO_3^-] + K_(a_1)K_(a_2)[HCO_3^(-)] = 1.0*10^-5 * K_(a_1) [H^+]=>[H+]2[HCO−3]+Ka1[H+][HCO−3]+Ka1Ka2[HCO−3]=1.0⋅10−5⋅Ka1[H+]⇒

[HCO_3^(-)]([H^+]^2 + K_(a_1)[H^+] + K_(a_1)K_(a_2))= 1.0*10^-5 * K_(a_1) [H^+][HCO−3]([H+]2+Ka1[H+]+Ka1Ka2)=1.0⋅10−5⋅Ka1[H+]

therefore [HCO_3^(-)] approx 2.5*10^-6M

From here, it's a lot easier! We'll use the former equilibrium expressions,

[H_2CO_3] = ((1.3*10^-6)(2.5*10^-6))/(4.3*10^-7) approx 7.5*10^-6, and

[CO_3^(2-)] = ((5.6*10^-11)(2.5*10^-6))/(1.3*10^-6) approx 1.1*10^-10

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