KVPY SA Exam 2018 Question Paper

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Question : 33 of 80

Marks: +1, -0

The solubility of BaSO

_4

in pure water (in g L

^{–1}

) is closest to
[Given : Ksp for BaSO

_4

is 1.0 × 10

^{–10}

at 25 ºC. Molecular weight of BaSO

_4

is 233 g mol

^{–1}

]

1.0 × 10 $^{–5 }$
1.0 × 10 $^{–3}$
2.3 × 10 $^{–5 }$
2.3 × 10 $^{–3}$

Solution:

Given Ksp = 10

^{–10}

For BaSO

^{4}

, Ksp = S

^{2}

.
S=10

^{–5 }

mol/L ⇒ 2.33 ×10

^{–3}

g/L

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