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.
I’m pretty sure it’s A it’s been awhile
