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Question

In 1L saturated solution of A g C l blank left square bracket K subscript s p end subscript open parentheses A g C l close parentheses 1.6 blank 10 to the power of 10 end exponent right square bracket, 0.1 mole of C u C l blank left square bracket K subscript s p end subscript open parentheses C u C l close parentheses 1.0 blank 10 to the power of 6 end exponent right square bracket is added. The resultant concentration of Ag in the solution is 1.6 blank 10 to the power of x end exponent. The value of ´ x ´ is

  1. 3  
  2. 5  
  3. 7  
  4. 9  

The correct answer is: 7


    It is a case of simultaneous solubility of salts with a common ion. Here solubility product of C u C l blankis much greater than that of A g C l, it can be assumed that C l to the power of minus end exponent in solution comes mainly from C u C l.
    rightwards double arrow open square brackets C l to the power of minus end exponent close square brackets equals square root of K subscript s p end subscript left parenthesis C u C l right parenthesis end root equals 10 to the power of negative 3 end exponent blank M
    Now for A g C l blank colon K subscript s p end subscript equals 1.6 cross times 10 to the power of negative 10 end exponent
    equals open square brackets A g to the power of plus end exponent close square brackets left square bracket C l to the power of minus end exponent right square bracket
    equals open square brackets A g to the power of plus end exponent close square brackets cross times 10 to the power of negative 3 end exponent
    rightwards double arrow open square brackets A g to the power of plus end exponent close square brackets equals 1.6 cross times 10 to the power of negative 7 end exponent

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