Question

The solubility of Baso, in water is 2.42* 10^3 L-
298 K. The value of its solubility product (K) wi
(Given molar mass of Baso, = 233 g mol-1)
(NEET-2018
(1) 1.08 * 10-10 mol-2
(2) 1.08 * 10-12 mol-2
(3) 1.08 * 10-mol
(4) 1.08 * 10-14 mol AL-2​

Answer 1

Answer...............

Option A

Answer 2

Answer :-

• $\sf\red{1.08\times 10^{-8}}$

Explanation:

Solubility product is defined as the equilibrium constant in which a solid ionic compound is dissolved to produce its ions in solution.

It is represented as $\blue{K_{sp}}$

The equation for the ionization of $BaSO_4$ is given as:

$\sf \blue{BaSO_4 \implies \: Ba^{2+}+SO_4^{2-}} \: \: { \bigstar}$

$</p><p>\bold{\underline{We \: are \: given:}}$

$\sf{ Solubility \: of \: BaSO_4}$

$\sf {</p><p> </p><p> \implies 2.42\times 10^{-3}}$

$\sf{</p><p></p><p>Molar \: mass \: of \: BaSO_4}</p><p>$

$\sf{</p><p> \implies 233 g/mol}$

$\sf{</p><p></p><p>Molar \: solubility \: of \: BaSO_4}$

$\sf{</p><p></p><p> \implies \dfrac{2.42\times 10^{-3}g/L}{233g/mol}=10.38\times 10^{-3}mol/L }</p><p>$

$\bold{</p><p>By \: stoichiometry \: of \: the \: reaction:}$

$\sf{\implies </p><p></p><p>1 \:mole \:of \: BaSO_4. \:</p><p></p><p> gives \: 1 \: mole \: of \:Ba^{2+} \:</p><p> and \: 1 \:mole \:of \:SO_4^{2-}</p><p> }$

$\bold{</p><p>Expression \: for \: the \: equilibrium \: constant \: of \: BaSO_4 \: will \: be \: :}$

$\sf{</p><p></p><p> \implies K_{sp}=[Ba^{2+}]^2[SO_4^{2-}]}$

$\sf \green{</p><p> </p><p>\implies K_{sp}=[10.38\times 10^{-3}][10.38\times 10^{-3}]=1.08\times 10^{-8}} \: \: { \bigstar}$

$\underline{ \underline{ \frak {\red{</p><p>Hence, \: the \: solubility \: product \:of \:BaSO_4</p><p> \: is \: 1.08\times 10^{-8}</p><p> }}}}</p><p>$