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Equilibrium

Question
CBSEENCH11006604

The solubility of Ag2CrO4 at 298K is 6·6 × 10–5 moles/litre. Find its solubility product.

Solution
We know,
Ag subscript 2 CrO subscript 4 space space space rightwards harpoon over leftwards harpoon space space space space 2 Ag to the power of plus space plus space CrO subscript 4 superscript 2 minus end superscript
Now 1 mole of  on dissociation gives 2 moles of Ag+ ions and 1 mole of CrO subscript 4 superscript 2 minus end superscript ions. 
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 ions.
Hence comma space
open square brackets Ag to the power of plus close square brackets space equals space 2 space cross times space 6.6 space cross times space 10 to the power of negative 5 end exponent mol space straight L to the power of negative 1 end exponent
open square brackets CrO subscript 4 superscript 2 minus end superscript close square brackets space equals space 6.6 space cross times 10 to the power of negative 5 end exponent space mol space straight L to the power of negative 1 end exponent
Applying space the space law space of space solubility space product comma 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 2 minus end superscript close square brackets
space space space space space space equals space left parenthesis 2 space cross times space 6.6 space cross times space 10 to the power of negative 5 end exponent right parenthesis squared space left parenthesis 6.6 space cross times space 10 to the power of negative 5 end exponent right parenthesis
space space space space space space equals 1.149 space cross times space 10 to the power of negative 12 end exponent
space space