SATHEE: Equilibrium - Result Question 70

Equilibrium - Result Question 70

####74. The $K_{s p}$ of $A g_{2} C r O_{4}, A g C l, A g B r$ and $A g I$ 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 $A g N O_{3}$ solution is added to the solution containing equal moles of $N a C l, N a B r, N a I$ and $N a_{2} C r O_{4}$ ? [2015]

(a) $A g C l$

(b) $A g B r$

(d) $A g I$

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Solution:

$K_{s p} = [A g^{+}]^{2} [C r O_{4}^{- 2}] = 1.1 \times 10^{- 12}$

$[A g^{+}] = \sqrt{\frac{1.1 \times 10^{- 12}}{[C r O_{4}^{- 2}]}}$

$A g C l$

$K_{s p} = [A g^{+}] [C l^{-}] = 1.8 \times 10^{- 10}$

$[A g^{+}] = \frac{1.8 \times 10^{- 10}}{[C l^{-}]}$

$A g B r$

$K_{s p} = [A g^{+}] [B r^{-}] = 5.0 \times 10^{- 13}$

$[A g^{+}] = \frac{5.3 \times 10^{- 13}}{[B r^{-}]}$

$A g I$

$K_{s p} = [A g^{+}] [I^{-}] = 8.3 \times 10^{- 17}$

$[A g^{+}] = \frac{8.3 \times 10^{- 17}}{[I^{-}]}$

If we take $[C r O_{4}^{- 2}] = [C l^{-}] = [B r^{-}] = [I^{-}] = 1$

then maximum $[A g^{+}]$ will be required in case of $A g_{2} C r O_{4}$ .

Equilibrium - Result Question 70

Solution:

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Equilibrium - Result Question 70

Solution:

Engineering Preparation

Medical Preparation

NCERT Books & Solutions

Important Resources

Other Resources

Syllabus

Test Platform

Sample Mock Test

Mentorship

NCERT Exercise Video Solution

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