The solubility product of PbBr 2 is 10 . 8 × 10 - 5 . It is 70 % dissociated in saturated solution. Calculate the solubility of the salt.

Let S be the solubility of the salt Degree of dissociation of the salt=0.7 PbBr 2 ⇌ Pb 2 + 0 . 7 S + 2 Br − 2 × 0 . 7 S K sp = [ Pb 2 + ] [ Br - ] 2 = ( 0 . 7 S ) × ( 1 . 4 S ) 2 = 0 . 98 S 3 or S 3 = 10 . 8 × 10 - 5 0 . 98 = 11 . 02 × 10 - 5 or S = 11 . 02 × 10 - 5 3 = 2 . 23 × 10 - 2 mol L - 1 Mol. mass of PbBr 2 = 367 Solubility of PbBr 2 = 2 . 23 × 10 - 2 × 367 = 8 . 2 g L - 1

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ALL IN ONE ISC Chemistry CLASS 11>Chapter 7 - Equilibrium>Chapter Practice>Q 30

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What is common ion effect?

Important Questions on Equilibrium

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ALL IN ONE ISC Chemistry CLASS 11>Chapter 7 - Equilibrium>Chapter Practice>Q 31

Define buffer solution. Give the application of buffer in daily life.

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ALL IN ONE ISC Chemistry CLASS 11>Chapter 7 - Equilibrium>Chapter Practice>Q 31

The value of $K_{c}$ in the reaction, $2 A ⇌ B + C$ is $2 \times 10^{- 3}$ . At a given time, the composition of reaction mixture is $A = B = C = 3 \times 10^{- 4} M$ . In which direction the reaction will proceed?