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Question: The conductivity of a saturated solution of \(\text{BaS}\text{O}_{4}\text{ is 3.06 }\text{×}\text{ ...

The conductivity of a saturated solution of

BaSO4 is 3.06 × 10-6 ohm-1 cm-1\text{BaS}\text{O}_{4}\text{ is 3.06 }\text{×}\text{ 1}\text{0}^{\text{-6}}\text{ oh}\text{m}^{\text{-1}}\text{ c}\text{m}^{\text{-1}}and its equivalent

conductance is 1.53 ohm-1 cm2 equiv-1.\text{1.53 oh}\text{m}^{\text{-1}}\text{ c}\text{m}^{2}\text{ equi}\text{v}^{\text{-1}}. The Ksp for

BaSO4 will be

A

4 × 10–12

B

2.5 × 10–13

C

25 × 10–9

D

10–6

Answer

10–6

Explanation

Solution

1.53 = 1000×3.06×10−6 Normality \frac { 1000 \times 3.06 \times 10 ^ { - 6 } } { \text { Normality } }

Normality = 2 × 10–3 M Molarity =2×10−32\frac { 2 \times 10 ^ { - 3 } } { 2 } = 10–3 M

Ksp = 10–6 M