Question

15. The solubility of PbSO4 at 25°C is 1.1 x 10-* mol/L. Then its Ksp is: *
A) 1.21 x 10-3
O B) 12.1 X 10-6
C) 121 x 10-11
O D) 1.21 x 10-10​

Answer 1

Explanation:

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Answer 2

Answer: $(1.21*10^{-16} )$

Explanation:

• Given, the solubility of $PbSO_{4}$ is $1.1*10^{-8} mol/l$.
• Temperature= 25°C
• We have to find $K_{sp}$ value.
• We know,

               $PbSO_{4}$ ↔ $Pb^{2+} + (SO_{4}) ^{2-}$

Let the solubility is taken as S for each ion.

                  s        ↔ s   +    s

             $K_{sp}$ = $S^{2}$                         where S= $1.1*10^{-8} mol/l$.

         ⇒  $K_{sp}$ =$(1.1*10^{-8} )^{2}$

         ∴   $K_{sp}$ = $(1.21*10^{-16} )$

• Hence, the required answer is  $(1.21*10^{-16} )$.

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