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Equilibrium

Question
CBSEENCH11006626
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The solubility product constant of Ag2CrO4 and AgBr are 1·1 × 10–12 and 5·0 × 10–13 respectively. Calculate the ratio of the molarities of their saturated solutions.

Solution
For Ag2CrO4, the solubility equilibrium is
Ag subscript 2 CrO subscript 4 space space rightwards harpoon over leftwards harpoon space space space 2 Ag to the power of plus left parenthesis aq right parenthesis space plus space CrO subscript 4 superscript 2 minus end superscript left parenthesis aq right parenthesis Let space the space solubility space of space Ag subscript 2 CrO subscript 4 space be space straight S space mol space straight L to the power of negative 1 end exponent therefore space space open square brackets Ag to the power of plus close square brackets space equals space 2 straight S semicolon space space space space open square brackets CrO subscript 4 superscript 2 minus end superscript close square brackets space equals space straight S space space space space space space straight K subscript sp space equals space open square brackets Ag to the power of plus close square brackets squared space open square brackets CrO subscript 4 superscript minus close square brackets space equals space left parenthesis 2 straight S right parenthesis squared left parenthesis straight S right parenthesis space equals space 4 straight S cubed or space space space space 4 straight S cubed space equals space straight K subscript sp space space or space space space straight S cubed space equals space straight K subscript sp over 4 space space 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 therefore space space space space straight S space equals space open parentheses fraction numerator 1.1 space cross times space 10 to the power of negative 12 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 space 0.65 space cross times space 10 to the power of negative 4 end exponent mol space straight L to the power of negative 1 end exponent
Thus,   straight M subscript Ag subscript 2 CrO subscript 4 end subscript space equals space 0.65 space cross times space 10 to the power of negative 4 end exponent space mol space straight L to the power of negative 1 end exponent
For AgBr, the solubility equilibrium is
 AgBr space space rightwards harpoon over leftwards harpoon space space space Ag to the power of plus left parenthesis aq right parenthesis space plus space Br to the power of minus left parenthesis aq right parenthesis
Let the solubility of AgBr be S mol L-1

therefore space space space space open square brackets Ag to the power of plus close square brackets space equals space straight S space semicolon space space open square brackets Br to the power of minus close square brackets space equals space straight S space space space space space space space space space space straight K subscript sp space equals space open square brackets Ag to the power of plus close square brackets space open square brackets Br to the power of minus close square brackets space equals space left parenthesis straight S right parenthesis space left parenthesis straight S right parenthesis space equals straight S squared or space space space space space space space space straight S squared space equals space straight K subscript sp space space end subscript or space space space straight S space equals space left parenthesis straight K subscript sp right parenthesis to the power of 1 divided by 2 end exponent or space space space space space space space space space straight S space equals space left parenthesis 5.0 space cross times 10 to the power of negative 13 end exponent right parenthesis to the power of 1 divided by 2 end exponent space space space space space space space space space space space space space space space equals space 0.707 space cross times space 10 to the power of negative 6 end exponent space mol space straight L to the power of negative 1 end exponent Thus space space space straight M subscript AgBr space equals space 0.707 space cross times space 10 to the power of negative 6 end exponent space mol space straight L to the power of negative 1 end exponent
The ratio of molarities of their saturated solutions
      equals space straight M subscript Ag subscript 2 CrO subscript 4 end subscript over straight M subscript AgBr space equals space fraction numerator 0.65 space cross times space 10 to the power of negative 4 end exponent over denominator 0.707 space cross times space 10 to the power of negative 6 end exponent end fraction space equals space 91.9

Since the solubility of Ag2CrO4 is more than that of AgBr, so the former is more soluble.
 

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?

What type of equilibrium is attained in a reversibe chemical change?

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