SATHEE: Equilibrium - Result Question 74

Equilibrium - Result Question 74

####78. The values of $K_{s p}$ of $C a C O_{3}$ and $C a C_{2} O_{4}$ are $4.7 \times 10^{- 9}$ and $1.3 \times 10^{- 9}$ respectively at $25^{\circ} C$ . If the mixture of these two is washed with water, what is the concentration of $C a^{2 +}$ ions in water?

[NEET Kar. 2013]

(a) $7.746 \times 10^{- 5} M$

(b) $5.831 \times 10^{- 5} M$

(d) $3.606 \times 10^{- 5} M$

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Solution:

(a) $C a C O_{3} ⟶ \underset{x}{C a^{2 +}} + C O_{3}^{2 -}$

$C a C_{2} O_{4} ⟶ \underset{y}{C a^{2 +}} + \underset{y}{C_{2} O_{4}^{2 -}}$

$∴ [C a^{2 +}] = x + y$

Now, $K_{s p} (C a C O_{3}) = [C a^{2 +}] [C O_{3}^{2 -}]$

or $4.7 \times 10^{- 9} = (x + y) x$

similarly, $K_{s p} (C a C_{2} O_{4}) = [C a^{2 +}] [C_{2} O_{4}^{2 -}]$

or $1.3 \times 10^{- 9} = (x + y) y$

On solving, we get

$[C a^{2 +}] = 7.746 \times 10^{- 5} M$

Equilibrium - Result Question 74

Solution:

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Equilibrium - Result Question 74

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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