Question

Calculate the concentration of all species in a 0.540 M solution of H2SO3. The acid ionization constants for the acid are Ka1=1.6×10^−2 and Ka2=6.4×10^−8.

Answer 1

"H_2 SO_3 is a diprofic acid New one can calculate p/t as follows: - K_a = [H_3 o^+] [HSO^- _3]/H_2 SO_3] But ka_1 = 1.6 times 10^-2 doublerightarrow 1.6 times 10^-2 = (x) (x)/(0.540 - x) doublerightarrow -x^2 - 1.6 times 10^-2 (0.540 - x) = 0 doublerightarrow -x^2 - 8.64 times 10^-3 + 1.6 times 10^-2 x = 0 on solving the above quadrafic equation x = 0.12 therefore [H_2 SO_3] = 0.540 -x = 0.540 - 0.122 = 0.418 Therefore [H_2 SO_3] = 0.540 - x = 0.540 - 0.122 = 0.418 [HSO_3] = 0.122, [H_3 O^+] = 0.122 K_92 = [H_3 O^+] [SO^2- _3]/[HSO^- _3] k_92 = 6.4 times 10^-8 6.4 times 10^-8 = (0.122 - x) (x)/(0.122) therefore x = 6.4 times 10^-8 therefore [H_3 O^+] = 0.122 m + 6.4 times 10^-8 m = 0.122 m [OH^-] = 1.0 times 10^-14