Question

The solubility product of BaCl2 is 4.0 x 10 –8

what will be its molar solubility

in mol dm -3

? Ans: ( S = 1x10-2 mol dm-3)​

Answer 1

Explanation:

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Answer 2

Given:

The solubility product of BaCl₂ is

$4.0 \times {10}^{ - 8}$

To Find:

The molar solubility.

Solution:

To find the molar solubility of BaCl₂, we will follow the following steps:

As we know,

BaCl₂

$Bacl₂ → {Ba}^{2 + } + {2cl}^{ - }$

Let molar solubility be s.

Then Solubility will be:

$s \: and \: 2s \: for \: {Ba}^{2 + } and \: {cl}^{ - } respectively$

Now,

Ksp = solubility product

$ksp = [{Ba}^{2 + }] {[ {cl}^{ - } ]}^{2}$

So,

$ksp = s \times {2s}^{2} = 4 {s}^{3}$

$4.0 \times {10}^{ - 8} = 4{s}^{3}$

$s = \sqrt[3]{ \frac{4.0 \times {10}^{ - 8} }{4} } = \sqrt[3]{10 \times {10}^{ - 9} } = 2.15 \times {10}^{ - 3} mol {dm}^{ - 3} = 0.2 \times {10}^{ - 2} mol {dm}^{ - 3} \: = 1 \times {10}^{ - 2} mol {dm}^{ - 3}$

Henceforth, molar solubility is

$1 \times {10}^{ - 2} mol {dm}^{ - 3}$

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