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Some carbonate compounds have limited solubility.MCO3(s)   ⇋  M2+(aq)  +  CO32-(aq)The solubility constant Ksp for a carbonate compound MCO3, Ksp = 7.2E-7.A saturated solution of the compound MCO3, is made.Calculate the concentration of the ions M2+(aq)  and  CO32-(aq) in the saturated solution (check the stoichiometry, the ions will have the same concentration).Give your answer to three significant figures, using the exponential format, for example  enter 0.0000382  as 3.82e-5 with no gaps and the -ve e if power is -ve.

Question

Some carbonate compounds have limited solubility.MCO3(s)   ⇋  M2+(aq)  +  CO32-(aq)The solubility constant Ksp for a carbonate compound MCO3, Ksp = 7.2E-7.A saturated solution of the compound MCO3, is made.Calculate the concentration of the ions M2+(aq)  and  CO32-(aq) in the saturated solution (check the stoichiometry, the ions will have the same concentration).Give your answer to three significant figures, using the exponential format, for example  enter 0.0000382  as 3.82e-5 with no gaps and the -ve e if power is -ve.

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Solution

The solubility product constant expression for the reaction is:

Ksp = [M2+][CO32-]

Given that the stoichiometry of the reaction is 1:1, the concentration of M2+ and CO32- ions in the solution will be the same. Let's denote this concentration as x. Therefore, we can rewrite the expression as:

Ksp = x^2

We know that Ksp = 7.2E-7. So, we can solve for x:

x^2 = 7.2E-7 x = sqrt(7.2E-7) x = 8.485E-4

Therefore, the concentration of M2+ and CO32- ions in the solution is 8.485E-4 M.

This problem has been solved

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