Question

Calculate the molar concentration of H+, HCO3-, and CO3-2 in a 0.200M Na2CO3 solution adjusted to pH 8.10.
Ka1 = 4.4 x 10^-7
Ka2 = 4.7 x 10^-11

Answer 1

Ka1 = ${\frac {[H^+][HCO_3^-]} {[H_2CO_3]}}$ = 4.4*10^-7

Ka2 = ${\frac {[H^+][CO_3^{2-}]} {[HCO_3^-]}}$ = 4.7*10^-11

Na₂CO₃ -> 2Na⁺ + CO₃^2-

Ka1*Ka2 = ${\frac {[H^+][HCO_3^-]} {[H_2CO_3]}}* {\frac {[H^+][CO_3^{2-}]} {[HCO_3^{-}]}}$ =${\frac {[H^+]^2[CO_3^{2-}]} {[H_2CO_3]}}=4.4*10^{-7}*4.7*10^{-11}=2.068*10^{-17}$

[H⁺] = 10^(-pH) = 10^(-8.10) = 7.943*10^-9

${\frac {[H^+]^2[CO_3^{2-}]} {[H_2CO_3]}}={\frac {(7.943*10^{-9})^2[CO_3^{2-}]} {[H_2CO_3]}}=2.068*10^{-17}$

${\frac {[CO_3^{2-}]} {[H_2CO_3]}}= {\frac {2.068*10^{-17}} {6.309*10^{-17}} }=0.3278$

[CO₃^2-] = x then [H₂CO₃] = 3.051x

the total amount of [CO₃^2-], [HCO₃^-] and [H₂CO₃] is 0.2 M. Then [HCO₃^-] is 0.2 - x - 3.051x = 0.2-4.051x

Ka1 =${\frac {[H^+][HCO_3^-]} {[H_2CO_3]}}={\frac {(7.943*10^{-9})(0.2-4.051x)} {3.051x}}=4.4*10^{-7}$

Solving the equation we get

x = 0.001156 M = 1.156 * 10^-3 M = [CO₃^2-]

0.2-4.051x = 0.1953 M = [HCO₃^-]

The answer:

[H⁺] = 7.943 * 10^-9 M

[HCO₃^-] = 1.953 * 10^-1 M

[CO₃^2-] = 1.156 * 10^-3 M