Ionic equilibrium

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Question

K_sp of PbBr₂ (Molar mass= 367 ) is 3 .2 x 10^-5 • If the salt is 80% dissociated in solution, calculate the solubility of salt in gram per litre. (in litre^-1)

Moderate

Solution

Let solubility of PbBr₂ is S mol litre^-1
$P b B r_{2} (s) + a q ⇌_{\frac{S \times 80}{100}} P b^{2 +} + \underset{\frac{2 s \times 80}{100}}{2 B r^{-}}$
Since, PbBr₂ ionizes to 80% only.
Now, $K_{s p} = [P b^{2 +}] {[B r^{-}]}^{2}$
or $\begin{array}{l} 3.2 \times 10^{- 5} = [\frac{S \times 80}{100}] {[\frac{2 S \times 80}{100}]}^{2} \\ S = 0.025 mollitre e^{- 1} \\ = 0.025 \times 367 {glitre}^{- 1} \\ = 9.175 {glitre}^{- 1} \end{array}$

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Ionic equilibrium

Unlock the full solution & master the concept.

Ksp of PbBr2 (Molar mass= 367 ) is 3 .2 x 10-5 • If the salt is 80% dissociated in solution, calculate the solubility of salt in gram per litre. (in litre-1)

Let solubility of PbBr2 is S mol litre-1 PbBr2(s)+aq⇌S×80100Pb2++2Br−2s×80100Since, PbBr2 ionizes to 80% only. Now, Ksp=Pb2+Br−2or 3.2×10−5=S×801002S×801002S=0.025 mollitre e−1=0.025×367glitre−1=9.175glitre−1

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K_sp of PbBr₂ (Molar mass= 367 ) is 3 .2 x 10^-5 • If the salt is 80% dissociated in solution, calculate the solubility of salt in gram per litre. (in litre^-1)