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The solubility of BaSO $_4$ in water is $2.42 \times 10^{-3}$ g/L at 298 K. The value of its solubility product (K $_{sp})$ will be $\\$ (Given molar mass of BaSO $_4 = 233$ g/mol)

medium

Equilibrium

2018

chemistry

$1.08 \times 10^{-14}$ mol $^2$ L $^{-2}$

Explanation

To find the solubility product $(K_{sp})$ of BaSO $_4,$ we need to follow these steps: $\\$

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$1.08 \times 10^{-12}$ mol $^2$ L $^{-2}$

$1.08 \times 10^{-10}$ mol $^2$ L $^{-2}$

$1.08 \times 10^{-8}$ mol $^2$ L $^{-2}$

1. Calculate the molar solubility of

BaSO

_4: \\

Given solubility of

BaSO

_4

in water is

2.42 \times 10^{-3}

g/L.

\\

Molar mass of

BaSO

_4 = 233

g/mol.

\\

Molar solubility

(s)

is given by:

\\ s = \frac{Solubility \ in \ g/L}{Molar \ mass \ in \ g/mol} \\ s = \frac{2.42 \times 10^{-3}}{233}

mol/L

\\

Calculate

s: \\ s = 1.039 \times 10^{-5}

mol/L

\\

2. Write the dissociation equation for

BaSO

_4: \\

BaSO

_4 (s) \rightleftharpoons

Ba

^{2+} (aq) +

SO

_4^{2-} (aq) \\

At equilibrium, the concentration of

Ba

^{2+}

and

SO

_4^{2-}

is equal to

s. \\

3. Write the expression for the solubility product

(K_{sp}): \\ K_{sp} = [

Ba

^{2+}][

SO

_4^{2-}] \\

Since both ions have the same concentration

s,

we have:

\\ K_{sp} = s \times s = s^2 \\

4. Substitute the value of

s

into the

K_{sp}

expression:

\\ K_{sp} = (1.039 \times 10^{-5})^2 \\ K_{sp} = 1.08 \times 10^{-10}

mol

^2

L

^{-2} \\

Therefore, the solubility product

K_{sp}

of

BaSO

_4

is

1.08 \times 10^{-10}

mol

^2

L

^{-2}. \\

This corresponds to Option 3.