Equilibrium

Question

At 298K, the, solubility of silver chloride in water is 0·00188 gL^–. What is its K_sp?

Solution

Molecular mass of AgCl
= 108 + 35.5 = 143.5
Solubility of AgCl in water = 0.00188 gL^-1

equals space fraction numerator 0.00189 over denominator 143 end fraction space mol space straight L to the power of negative 1 end exponent

equals space 1.31 space cross times space 10 to the power of negative 5 end exponent space mol space straight L to the power of negative 1 end exponent

We Know,

AgCl space space rightwards harpoon over leftwards harpoon space space space Ag to the power of plus space plus space Cl to the power of minus

i.e. 1 mole of AgCl in solution gives 1 mole of Ag⁺ ion and 1 mole of Cl^-ions. Hence in the solution

open square brackets Ag to the power of plus close square brackets space equals space 1.31 space cross times space 10 to the power of negative 5 end exponent space mol space straight L to the power of negative 1 end exponent open square brackets Cl to the power of minus close square brackets space equals space 1.31 space cross times space 10 to the power of negative 5 end exponent space mol space straight L to the power of negative 1 end exponent

Applying the law of solubility product,

straight K subscript sp space equals space open square brackets Ag to the power of plus close square brackets space open square brackets Cl to the power of minus close square brackets space space space space space space equals left parenthesis 1.31 space cross times space 10 to the power of negative 5 end exponent right parenthesis thin space left parenthesis 1.31 space cross times space 10 to the power of negative 5 end exponent right parenthesis space space space space space space equals space 1.7 space cross times space 10 to the power of negative 10 end exponent

Some More Questions From Equilibrium Chapter

For Daily Free Study Material Join wiredfaculty