A sparingly soluble salt gets precipitated only when the product of concentration of its ions in the solution (Q_sp) becomes greater than its solubility product. If the solubility of BaSO₄ in water is 8 × 10^–4 mol dm^–3. Calculate its solubility in 0.01 mol dm^–3 of H₂SO₄.

<p>This is a Short Answer Type Questions as classified in NCERT Exemplar</p><p><strong>Ans: </strong>The chemical equation is given below-&nbsp;&nbsp;</p><p>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; BaSO4(s) → Ba2+(aq)&nbsp;+ SO42-(aq)</p><p>At t = 0 &nbsp; &nbsp; &nbsp; &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; 1 &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; 0 &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; 0&nbsp;</p><p>At equilibrium &nbsp;in water &nbsp; &nbsp; &nbsp; &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; 1-S &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; S &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; S&nbsp;</p><p>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;</p><p>At equilibrium in&nbsp;the presence &nbsp; &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; 1-S &nbsp; &nbsp; &nbsp; &nbsp; &nbsp;&nbsp;&nbsp; S&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp;&nbsp;&nbsp;&nbsp;&nbsp; (S+0.01)&nbsp;</p><p>of sulphuric acid&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;</p><p>K_sp&nbsp;for BaSO₄&nbsp;in water =&nbsp;[&nbsp;Ba₂⁺&nbsp;]&nbsp;[&nbsp;SO₂^4-&nbsp;]&nbsp;=&nbsp;(S)&nbsp;(S)&nbsp;=&nbsp;S₂</p><p>But S = 8 x 10^-4&nbsp;mol dm-3</p><p>K_sp&nbsp;=&nbsp;(8&nbsp;×&nbsp;10^-4)2&nbsp;= 64&nbsp;×&nbsp;10^-8&nbsp;. (1)&nbsp;</p><p>The expression for K_sp in the presence of sulphuric acid will be as follows:&nbsp;</p><p>K_sp = (S) (5 + 0.01) . (2)&nbsp;</p><p>Since value of K_sp will not change in the presence of sulphuric acid, therefore from (1) and (2)&nbsp;</p><p>(S) (S + 0.01) = 64&nbsp;×&nbsp;10^-8</p><p>S² + 0.01 S = 64&nbsp;×&nbsp;10^-8</p><p>S² + 0.01 S - 64&nbsp;×&nbsp;10^-8= 0</p><p>S =<span contenteditable="false"> <math> <mfrac> <mrow> <mrow> <mo>-</mo> <mn>0.01</mn> <mo>±</mo> <mroot> <mrow> <mrow> <msup> <mrow> <mrow> <mfenced separators="|"> <mrow> <mrow> <mn>0.01</mn> </mrow> </mrow> </mfenced> </mrow> </mrow> <mrow> <mrow> <mn>2</mn> </mrow> </mrow> </msup> </mrow> </mrow> <mrow></mrow> </mroot> <mo>+</mo> <mi mathvariant="normal"> </mi> <msup> <mrow> <mrow> <mfenced separators="|"> <mrow> <mrow> <mn>4</mn> <mo>×</mo> <mn>64</mn> <mi mathvariant="normal">&nbsp;</mi> <mo>×</mo> <mi mathvariant="normal">&nbsp;</mi> <mn>10</mn> <mo>-</mo> <mn>8</mn> </mrow> </mrow> </mfenced> </mrow> </mrow> <mrow> <mrow> <mn>2</mn> </mrow> </mrow> </msup> <mi mathvariant="normal">&nbsp;</mi> </mrow> </mrow> <mrow> <mrow> <mn>2</mn> </mrow> </mrow> </mfrac> </math> </span></p><p>When we solve the above equation, we will get the value of</p><p>&nbsp;S is= 6&nbsp;×&nbsp;10^-4&nbsp;mol^-1&nbsp;dm^-3</p>

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