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Equilibrium

Question
CBSEENCH11006612
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Determine the solubility of silver chromate at 298K. Given that Ksp for Ag2CrO4 = 1·1 × 10–12. Also determine the molarities of each ion. 

Solution
Let the solubility of Ag2CrO4 in water be S mol L–1.
Ag subscript 2 CrO subscript 4 left parenthesis straight s right parenthesis space space rightwards harpoon over leftwards harpoon space 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 straight i. straight e. space 1 space mole space of space Ag subscript 2 CrO subscript 4 space in space the space solution space gives space 2 space moles space of space Ag to the power of plus space ions space and space 1 space mole space of space CrO subscript 4 superscript 2 minus end superscript space ion.
Hence in solution, 
therefore space space open square brackets Ag to the power of plus close square brackets space equals space 2 straight S space mol space straight L to the power of negative 1 end exponent semicolon space space space space open square brackets CrO subscript 4 superscript 2 minus end superscript close square brackets space equals 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 Ag to the power of plus close square brackets squared space open square brackets CrO subscript 4 superscript 2 minus end superscript 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 straight S cubed space equals space straight K subscript sp over 4 or space 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 1.1 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 space space space space space equals space 0.65 space cross times 10 to the power of negative 4 end exponent space mol space straight L to the power of negative 1 end exponent therefore space space space Molarity space of space Ag to the power of plus space ions space space space open square brackets Ag to the power of plus close square brackets space equals space 2 straight S space equals space 2 cross times 0.65 space cross times space 10 to the power of negative 4 end exponent space space space space space space space space space space space space equals 1.03 space cross times space 10 to the power of negative 4 end exponent straight M
Molarity space fo space CrO subscript 4 superscript 2 minus end superscript space ions comma space space space space space space open square brackets CrO subscript 4 superscript 2 minus end superscript close square brackets space equals space straight S space equals space 0.65 space cross times space 10 to the power of negative 4 end exponent straight M

Some More Questions From Equilibrium Chapter

What kind of molecules in a liquid can evaporate?

Name the factors on which vapour pressure of any liquid depends.

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?

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