SATHEE: Equilibrium - Result Question 80

Equilibrium - Result Question 80

####84. In qualitative analysis, the metals of Group I can be separated from other ions by precipitating them as chloride salts. A solution initially contains $A g^{+}$ and $P b^{2 +}$ at a concentration of $0.10 M$ . Aqueous $H C l$ is added to this solution until the $C l^{-}$ concentration is $0.10 M$ . What will the concentrations of $A g^{+}$ and $P b^{2 +}$ be at equilibrium?

$(K_{s p} .$ for $A g C l = 1.8 \times 10^{- 10}$ ,

$K_{s p}$ for $P b C l_{2} = 1.7 \times 10^{- 5}$ )

[2011M]

(a) $[A g^{+}] = 1.8 \times 10^{- 7} M; [P b^{2 +}] = 1.7 \times 10^{- 6} M$

(b) $[A g^{+}] = 1.8 \times 10^{- 11} M; [P b^{2 +}] = 8.5 \times 10^{- 5} M$

(d) $[A g^{+}] = 1.8 \times 10^{- 11} M; [P b^{2 +}] = 8.5 \times 10^{- 4} M$

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Solution:

$1.8 \times 10^{- 10} = [A g^{+}] [0.1]$

$[A g^{+}] = 1.8 \times 10^{- 9} M$

$K_{s p} = [P b^{+ 2}] [C l^{-}]^{2}$

$1.7 \times 10^{- 5} = [P b^{+ 2}] [0.1]^{2}$

$[P b^{+ 2}] = 1.7 \times 10^{- 3} M$

Equilibrium - Result Question 80

Solution:

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Equilibrium - Result Question 80

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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