Answer given: Explanation with solution setup please Calculate the concentration

Question

Answer given:

Explanation with solution setup please

Calculate the concentrations of H3O+, OH- , HSeO4- , and SeO42- in 0.15 M H2SeO4, selenic acid, solution. [H3O+] = 0.16 M, [OH-] = 6.2 times 10-14 M, [HSeO4-] = 0.14 M, [SeO42-] = 0.01M

Explanation / Answer

H2SeO4 dissociates completely into HSeO4- and H+ So you have concentration of HSeO4 - and H+ is 0.15M Then HSeO4 - ====> SeO42- + H+ At equilibrium SeO42- = +x H+ = .15 + x HSeO4- = .15 - x Ka2 = 2.2 x 10^-2 = x(.15+x)/(.15-x) Solve for x, x = 0.017 M [H3O+] = .15 + .017 = 0.167 M [OH-] = 10^-14/.167 = 6 x 10^-14 M [HSeO4-] = .15 - .017 = .133 M [SeO42-] = .017 M Hope this helped

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