If the solubility of AgCl (formula mass=143) in water at 25oC is 1.43 ×10−4 g/100 mL of solution then the value of Ksp will be____________.
Given,
Formula Mass of AgCl=143
Solubility of AgCl=1.43×10−4g/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
AgCl→Ag++Cl−
Now, concentration of each of the ions can be given as
Concentration=1.43×10−4143mol/100mL
⇒10−6mol/100mL
Converting mL into Liter
⇒10−6×1000100mol/Liter
⇒10−5mol/L
Substituting this value in the formula to calculate Ksp we get,
Ksp=[Ag+]1[Cl−]1
Ksp=[10−5][10−5]
Ksp=10−10
Therefore the value of Ksp for AgCl is 1×10−10 .
Hence, option C is the correct option.