SATHEE: Equilibrium - Result Question 67

Equilibrium - Result Question 67

####70. The solubility of $B a S O_{4}$ in water is $2.42 \times 10^{- 3} g L^{- 1}$ at $298 K$ . The value of its solubility product $(K_{s p})$ will be

(Given molar mass of $B a S O_{4} = 233 g m o l^{- 1}$ )

(a) $1.08 \times 10^{- 10} m o l^{2} L^{- 2}$

[2018]

(b) $1.08 \times 10^{- 12} m o l^{2} L^{- 2}$

(d) $1.08 \times 10^{- 14} m o l^{2} L^{- 2}$

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Solution:

(a) Solubility of $B a S O_{4} = 2.42 \times 10^{- 3} g L^{- 1}$

$∴ s = \frac{2.42 \times 10^{- 3}}{233} = 1.038 \times 10^{- 5} m o l L^{- 1}$

$K_{s p} = s^{2} = (1.038 \times 10^{- 5})^{2} = 1.08 \times 10^{- 10} m o l^{2} L^{- 2}$

Equilibrium - Result Question 67

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Equilibrium - Result Question 67

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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