Question

Calculate the concentration of [H2CO3], [HCO3(-)], [CO3 (2-)], [H3O(+)], and [OH(-)] in a 0.140M solution of H2CO3. In paranthesis, are the charges of the ions.

Answer 1

we need to know the K_a1 and K_a2 values of H₂CO₃ . my values are K_a1=4.3 * 10^-7 and K_a2=5.6 * 10^-11

H₂CO₃ HCO₃^- + H⁺

here K_a1=4.3 * 10^-7 = [H⁺][HCO₃^-]/[H₂CO₃]

that implies 4.3 * 10^-7 = x²/0.14-x

as x is small we can write 0.14-x as 0.14

therefore after solving we get x = 2.45 * 10^-4 = [HCO₃^-] = [H⁺]

for second reaction

HCO₃^- H⁺ + CO₃^-2

here K_a2 = 5.6 * 10^-11 = [H⁺][CO₃^-2]/[HCO₃^-] = y²/2.45 * 10^-4 -y

after solving we get y = [H⁺] = [CO₃^-2] = 1.17 * 10^-7

now total [H⁺] =[H₃O⁺] = 1.17 * 10^-7 + 2.45 * 10^-4 = 2.45117 * 10^-4

we know that K_w = [H⁺][OH^-] = 10^-14

therefore [OH^-] = 10^-14/2.45117 * 10^-4 = 4.079 * 10^-11