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Question

If the solubility of AgCl (formula mass=143) in water at 25oC is 1.43 ×104 g/100 mL of solution then the value of Ksp will be____________.

A
1×105
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B
2×105
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C
1×1010
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D
2×1010.
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Solution

The correct option is D 1×1010

Given,

Formula Mass of AgCl=143

Solubility of AgCl=1.43×104g/100mL

We know that Kspof a salt is given as Ksp=[A+]a[B]b

Where A+ is the concentration of cation in aqueous solution and B is the concentration of anion in the aqueous solution.

The dissociation of salt AgCl can be given as

AgClAg++Cl

Now, concentration of each of the ions can be given as

Concentration=1.43×104143mol/100mL

106mol/100mL

Converting mL into Liter

106×1000100mol/Liter

105mol/L

Substituting this value in the formula to calculate Ksp we get,

Ksp=[Ag+]1[Cl]1

Ksp=[105][105]

Ksp=1010

Therefore the value of Ksp for AgCl is 1×1010 .

Hence, option C is the correct option.


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