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Equilibrium

Question
CBSEENCH11006615
Wired Faculty App

Determine the solubility of lead chloride at 298K. Given Ksp for PbCl2 = 1·6 × 10–5. Also, determine the molarity of each ion.

Solution
Let the solubility of lead chloride in water be S mol L-1
PbCI subscript 2 left parenthesis straight s right parenthesis space space space rightwards harpoon over leftwards harpoon space space space space Pb to the power of 2 plus end exponent left parenthesis aq right parenthesis space plus space 2 Cl to the power of minus left parenthesis aq right parenthesis
i.e. 1 mole of PbCI2 in the solution gives 1 mole of Pb2+ ions and 2 moles of Cl- ions. 
Hence in the solution,
  open square brackets Pb to the power of 2 plus end exponent close square brackets space equals space straight S space mol space straight L to the power of negative 1 end exponent semicolon space space space open square brackets Cl to the power of minus close square brackets space equals space 2 straight S space mol space straight L to the power of negative 1 end exponent
Applying the law of solubility product
  straight K subscript sp space equals space open square brackets Pb to the power of 2 plus end exponent close square brackets space open square brackets Cl to the power of minus close square brackets squared space equals space straight S space cross times space space left parenthesis 2 straight S right parenthesis squared space equals space 4 straight S cubed
or space space space straight S space equals space open parentheses straight K subscript sp over 4 close parentheses to the power of 1 divided by 3 end exponent space equals space open parentheses fraction numerator 1.6 space cross times space 10 to the power of negative 5 end exponent over denominator 4 end fraction close parentheses to the power of 1 divided by 3 end exponent space space space space space space space space equals open parentheses fraction numerator 16 space cross times space 10 to the power of negative 6 end exponent over denominator 4 end fraction close parentheses to the power of 1 divided by 3 end exponent space space space space space space space equals space left parenthesis 4 space cross times space 10 to the power of negative 6 end exponent right parenthesis to the power of 1 divided by 3 end exponent space space space space space space space space equals space 1.58 space cross times space 10 to the power of negative 2 end exponent

Molarity space of space Pb to the power of 2 plus end exponent space ions comma space open square brackets Pb to the power of 2 plus end exponent close square brackets space equals space straight S space space equals 1.58 space cross times space 10 to the power of negative 2 end exponent straight M Molarity space of space Cl to the power of minus space ions comma space space open square brackets Cl to the power of minus close square brackets space equals space 2 straight S space equals space 2 space cross times space 1.58 space cross times space 10 to the power of negative 2 end exponent straight M space equals space 3.16 space cross times space 10 to the power of negative 2 end exponent straight M

Some More Questions From Equilibrium Chapter

Which measurable property becomes constant in water vapour equilibrium?

Give an example from daily life in which there is gas solution equilibrium?

A crystal of common salt of a given mass is kept in its aqueous solution. After 24 hours, its mass remains the same. Is the crystal in equilibrium with the solution?

What is the role of forward and backward reactions at equilibrium of a reversible reaction?

What is a forward reaction?

What condition favours a reversible reaction?

What is a backward reaction?

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