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The solubility product of PbI2 is 8.0×10−9. The solubility of lead iodide in 0.1 molar solution of lead nitrate is x×10−6 mol/L. Given: √2=1.41. The value of x is

Answer» The solubility product of PbI2 is 8.0×109. The solubility of lead iodide in 0.1 molar solution of lead nitrate is x×106 mol/L. Given: 2=1.41. The value of x is


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