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  • Stuck here, help me understand: - Equilibrium - NEET

The K_{sp} \ of \ Ag_{2}CrO_{4}, Agcl, AgBr\ and\ AgI are respectively, 1.1\times 10^{-12}, 1.8\times 10^{-10}, 5.0\times 10^{-13}, 8.3\times 10^{-17}. Which one of the following salts will precipitate last if AgNO_{3} solution is added to the solution containing equal moles of NaCl, NaBr, Nal, and Na_{2}CrO_{4}?

  • Option 1)

    AgCl

  • Option 2)

    AgBr

  • Option 3)

    Ag_{2}CrO_{4}

  • Option 4)

    AgI

 

Answers (1)

 

General expression of solubility product -

M_{x}X_{y}\rightleftharpoons xM^{p+}(aq)+yX^{2-}(aq)


(x.p^{+}=y.\bar{q})

- wherein

Its solubility product is 
 

K_{sp}=[M^{p+}]^{x}\:[X^{2-}]^{y}

 

 Ksp - solubility

AgCrO4     1.1X10-12= 4 S3       S=3\sqrt{\frac{k_{sp}}{4}}=0.65 \times 10^{-4}

Agcl   1.8X10-10= S2            S=\sqrt{k_{sp}}= 1.34X10^{-5}  

AgBr     5X1013 = S2                 S=\sqrt{K_{sp}}=0.71 \times 10^{-6}

AgI        8.3 X10-17 =  S2           S=\sqrt{K_{sp}}=0.9 \times 10^{-8}

Thus Ag2Cr04  will be precipitated last.


Option 1)

AgCl

incorrect

Option 2)

AgBr

incorrect

Option 3)

Ag_{2}CrO_{4}

correct

Option 4)

AgI

incorrect

Posted by

Vakul

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