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If the solubility product of silver chloride is 1.8 × 10-10. What is the solubility of silver ion if concentration of Cl– is 0.01 molar. |
Answer» Ksp = [Ag+][Cl-] 1.8 x 10-10 = [Ag+]0.01 [Ag+] = \(\frac{1.8\times 10^{-10}}{0.01}\) = 180 x 10-10 = 1.8 x 10-8 mol/dm3 |
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