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Equilibrium

Question
CBSEENCH11006617
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The solubility product of BaSO4 is 1·5 × 10–9. Find solubility in:
(i) pure water    
(ii) 0·10 M BaCl2.

Solution
Let the solubility of BaSO4 be S mole per litre.
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Applying the law of solubility product, 
      straight K subscript sp space equals space open square brackets Ba to the power of 2 plus end exponent close square brackets space open square brackets SO subscript 4 superscript 2 minus end superscript close square brackets space equals space straight S space cross times space space straight S or space space straight S squared space equals space straight K subscript sp or space space space straight S space equals space square root of straight K subscript sp end root space equals space square root of 1.5 space cross times space 10 to the power of negative 9 end exponent end root space space space space space space space space space space space equals space 3.78 space cross times space 10 to the power of negative 5 end exponent space mol space straight L to the power of negative 1 end exponent

left parenthesis ii right parenthesis space Solubility space in space 0.10 straight M space BaCl subscript 2 colon open square brackets Ba to the power of 2 plus end exponent close square brackets space equals space 0.1 space plus space straight S semicolon space open square brackets SO subscript 4 superscript 2 minus end superscript close square brackets space equals space straight S space space space straight K subscript sp space equals space open square brackets Ba to the power of 2 plus end exponent close square brackets space open square brackets SO subscript 4 superscript 2 minus end superscript close square brackets space space space space space space space space equals space space left parenthesis 0.1 plus straight S right parenthesis space left parenthesis straight S right parenthesis space equals space 0.1 space straight S space plus space straight S squared straight S squared space being space the space product space of space two space small space quanitities space is space neglected. therefore space space space straight K subscript sp space equals space 0.15 space therefore space space space 0.1 space straight S space equals space 1.5 space cross times space 10 to the power of negative 9 end exponent or space space space space space straight S space equals space fraction numerator 1.5 space cross times space 10 to the power of negative 9 end exponent over denominator 0.1 end fraction space equals space 1.5 space cross times space 10 to the power of negative 8 end exponent space mol space straight L to the power of negative 1 end exponent

Some More Questions From Equilibrium Chapter

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?

What is meant by a system at dynamic equilibrium?

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