Calculate the molar solubility of pbco3 when the ksp of pbco3 is 1.5*10^-15 at 25°c.

First, write the dissociation equation for PbCO3 which is PbCO3 = Pb^(2+) + CO3^(2-). Then, write the Ksp equation which is Ksp = [Pb^(2+)][CO3^(2-)]. The equation suggests that there is 1:1 molar ratio between the Pb^(2+), CO3^(2-) and the dissolved PbCO3. Thus in equation form, we can represent them as x. The Ksp equation is then: 1.5*10^-15 = x^2. The molar solubility of PbCO3, x, is then equal to 3.87*10^-8 moles per liter.

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