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Question: How do you write the \[\;Ksp\] expression for lead chromate \[PbCr{O_4}\] and calculate its solubili...

How do you write the   Ksp\;Ksp expression for lead chromate PbCrO4PbCr{O_4} and calculate its solubility in mol/Lmol/L? Ksp=2.310Ksp = 2.310?

Explanation

Solution

Compose the chemical equations and equilibrium expressions speaking to solubility equilibria .The Complete equilibrium calculations including solubility, equilibrium expressions, and solute concentration.

Complete step by step answer:
The conservation of clinical lab blood tests, digging of ocean water for magnesium, detailing of over-the-counter prescriptions, for example, Milk of Magnesia and antacids, and treating the presence of hard water in your home's water supply are only a couple of the numerous tasks that include controlling the equilibrium between a marginally dissolvable ionic solid and an aqueous solution of its ions
In an saturated solution of lead chromate, the solid is in equilibrium with its ions:
PbCrO4PbCr{O_4} ⇌\rightleftharpoons Pb2+P{b^{2 + }} + CrO42−Cr{O_4}^{2 - }
For which-
K = $$$$P{b^{2 + }}$$$$ + CrO42−Cr{O_4}^{2 - } = 2.3×10−13mol2l−22.3 \times {10^{ - 13}}mo{l^2}{l^{ - 2}}

This is temperature dependent and the temperature should have been determined in the question
From the equilibrium, you can see thatPb2+P{b^{2 + }} == CrO42−Cr{O_4}^{2 - }
{\left[ {P{b^{2 + }}} \right]^2}$$$$ = 2.3×10−13mol2l−22.3 \times {10^{ - 13}}mo{l^2}{l^{ - 2}}
\left[ {P{b^{2 + }}} \right]$$$$ = 2.3×10−13\sqrt {2.3 \times {{10}^{ - 13}}} == 4.87×10−7mol/l4.87 \times {10^{ - 7}}mol/l
Since lead chromate is 11 molar as for lead ions, at that point this is additionally equivalent to the dissolvability at that specific temperature.

Note: The solid itself doesn't show up as a denominator in the expression since it isn't in a similar phase as the aqueous ions. As a rule, the solubility product consistent KspKsp , is the equilibrium constant for the solvency equilibrium of a slightly soluble ionic compound.
Like all equilibrium constants, the  Ksp\;Ksp is temperature-dependent, yet at a given temperature it remains generally constant. It likewise essential, that, much the same as any equilibrium, every particle concentration in the expression is raised to the force of its coefficient in the solubility condition