1)A saturated solution 250 mL of AgCH3COO was evaporated to dryness. Once dry the sample was found to contain 1.84 g AgCH3COO. Calculate the solubility product constant for AgCH3COO.
2)
The molar solubility of Pb₃(PO₄)₂ is 6.3x10-12. Find its Ksp.
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Answer 1 - The solubility product constant for AgCH3COO = 1.94 × 10-3 . Explanation -  Answer rarrWe know that[" mol "=(" Given mass ")/(" molar mass ")]Since, molar mass of Ag CH_(3)COOrarr166*91g//mol.mol=(1.84)/(166.91)=0.0110molAgCH_(3)COOConcentration =>(0.0110(mol))/(250(L)xx0^(-3)(L))=0.0441(mol)/(L)AgctbCO0{:[AgCH_(3)COO_((s)),⇌Ag^(+)(aq) +CH_(3)COO_((aq) )],[0.0441,0.0441],[k_(sp),=[Ag^(+)][CH_(3)COO^(-)]],[ ... See the full answer