Predominant chemical form and concentration of chromium

Question

How do I find the concentration and predominant chemical form of chromium in a solution at pH6 and at pH8 with a 10^-4 molar of chromic acid H2CrO4? H2CrO4 pKa1 is 0.20 and pKa2 is 6.5

Answer 1

Actually calculating the concentrations is a bear, so I am assuming that you're going to assume that you are in a regime with one of the forms as dominant.

You can write the Henderson-Hasselbalch for each equilibrium:

pK_A1 - pH = log([HCrO₄^-]/[H₂CrO₄]) or the ratio of concentrations = 10^(pH-pK_A1) = C₁

pK_A2 - pH = log([CrO₄^2-]/[HCrO₄^-]) or the ratio of concentrations = 10^(pH-pK_A2) = C₂

Finding the value of the ratios for the two conditions:

pH = 6 : C₁ = 6.31 x 10⁵ and C₂ = 0.316 (HCrO₄^-) dominant

pH = 8 C₁ = 6.31 x 10⁷ and C₂ = 3160 (CrO₄^-) dominant

To solve for the quantities of the predominant form:

pH = 6 let x be the amount of [HCrO₄^-] reacted from an initial concentration of 10^-4 molar (assume that all the H₂CO₄ turned to HCrO₄^-. 2nd eqn. becomes x/(10^-4-x) = .316 and solve for x, Concentration of bichromate ion will be 10^-4-x.

pH=8 you could estimate the chromate ion concentration as 10^-4 or go through the same calculation as for a pH of 6 with a new C₂.

I have to stop here as I have spent too long answering the question. Take care.