Lead chromate has been widely used as a paint pigment, although this usage has been curtailed by env — Physical Chemistry Chemistry Question

Because of hydrolysis of chromate, the common rule that the solubility of a salt is equal to the square root of its Ks is not valid. There would be various paths to a solution of this problem, but most would begin with the mass balance equation, which simply states that the sum of dissolved Cr concentrations in all species must equal the Pb²⁺ concentration.
[Pb²⁺] = [CrO₄ 2–] + [HCrO₄–] + 2 [Cr₂O₇ 2–]

This equation plus the equilibrium constant expressions for Ks, Ka₂ and KD constitute a set of four equations containing four unknown variables: [Pb²⁺], [CrO₄ 2–], [HCrO₄ –], and [Cr₂O₇ 2–]. (Note that H⁺ is specified by the problem and is not an unknown variable.) Because the number of equations equals the number of variables, these equations can be solved for the values of the variables (concentrations).
Using the equilibrium constant expressions, each term on the right side of the mass balance equation can be replaced with a term containing [Pb²⁺] as follows:

Inserting [H⁺] = 1.00 · 10–6 and the values of the equilibrium constants and rearranging gives:
[Pb²⁺]3 = [Pb²⁺] (2.82 ⋅ 10-13 + 8.44 ⋅ 10-13) + 4.98 ⋅ 10-23
By trying various values of [Pb²⁺] in this equation, it becomes apparent that the final term will be negligible. Therefore, [Pb²⁺]2 = 1.13 · 10–12, and
[Pb²⁺] = 1.06 · 10–6
Back substituting into the equilibrium constant expressions gives:
[CrO₄ 2–] = 2.66 · 10–7
[HCrO₄ –] = 7.96 · 10–7
[Cr₂O₇ 2–] = 2.21 · 10–11
The answers should be given to three significant figures.

In this case, we want to know the total dissolved Cr concentration c(Cr)T at which [HCrO₄ –] = [Cr₂O₇ 2–] at pH 3.00. The mass balance equation is:
[Cr] T = [CrO₄ 2–] + [HCrO₄–] + 2 [Cr₂O₇ 2–]
This equation also contains four unknown variables. The expressions for Ka₂ and KD, as well as this equation and the constraint that [HCrO₄ –] = [Cr₂O₇ 2–] give the four equations needed for solution. If we replace [CrO₄ 2–] using the Ka₂ expression, we obtain:
[CrT] = 3.34 · 10–7 [HCrO₄ –] / [H⁺] + [HCrO₄–] + 2 [Cr₂O₇ 2–]
At pH 3.00, it is clear that the first term is negligible relative to the second one. Furthermore, because we are interested in a situation where the third term is twice the second, it follows that the first term is negligible also relative to the third. If we drop the first term and let the concentrations of HCrO₄ – and Cr₂O₇ 2– be x, then
[CrT] = x + 2 x, and x = [CrT] / 3
To find [CrT], we make use of the expressions for KD and Ka₂:

So, [CrT] = 0.0859 or c(CrT) = 0.0859 mol dm–3.
Note that the answer should be quoted to three significant figures. It might be argued that for pH = 3.00 the [H⁺] concentration is known only to two figures, based on the rule that the number of significant figures in a logarithm is equal to the number of figures in the mantissa. However, in this problem the term in the mass balance equation that contains [H⁺] is negligible and [H⁺] cancels in the product, KD Ka2^2. Consequently the precision of [H⁺] does not affect the precision of the answer.
An alert chemist might have come to the correct answer a little more efficiently by inspecting the Ka₂ expression and observing that CrO₄ 2– must be negligible relative to HCrO₄ – at pH 3.00. On this basis, the alert chemist would have omitted [CrO₄ 2–] from the mass balance equation, saving a little time.