H2CrO4 ? 2 H+ + CrO42- Ka = 5.83 x 10-8 Initial .0375 2(.150) 0 Change - x +x +x Eq. .0375 -x .3 + x x Ka = [H+][A-]/[HA] 5.83E-8 = (.3 + x)(x)/(.0375 - x) Let\'s assume that the value of [H+] is within +- 5%, so we can simplify the above eq. to make calculations simpler. 5.83E-8 = (.3x + x^2)/(.0375) x = [CrO4^2-] = 7.3E-9 mol/L Solution H2CrO4 ? 2 H+ + CrO42- Ka = 5.83 x 10-8 Initial .0375 2(.150) 0 Change - x +x +x Eq. .0375 -x .3 + x x Ka = [H+][A-]/[HA] 5.83E-8 = (.3 + x)(x)/(.0375 - x) Let\'s assume that the value of [H+] is within +- 5%, so we can simplify the above eq. to make calculations simpler. 5.83E-8 = (.3x + x^2)/(.0375) x = [CrO4^2-] = 7.3E-9 mol/L.