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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 BaSO_(4) in water is 8xx10^(-4)mol dm^(-3) , calculate its solubility in 0.01 mol dm^(-3) of H_(2)SO_(4).

Answer» <html><body><p></p>Solution :Solubility of `BaSO_(4) ` in water `(S) = 8 xx 10^(-4) ` <a href="https://interviewquestions.tuteehub.com/tag/mol-1100196" style="font-weight:bold;" target="_blank" title="Click to know more about MOL">MOL</a> `dm^(-3)` <a href="https://interviewquestions.tuteehub.com/tag/ltbgt-2804401" style="font-weight:bold;" target="_blank" title="Click to know more about LTBGT">LTBGT</a> `BaSO_(4) hArr Ba^(<a href="https://interviewquestions.tuteehub.com/tag/2-283658" style="font-weight:bold;" target="_blank" title="Click to know more about 2">2</a>) + SO_(4)^(2-)` <br/> `:. K_(sp) ` for `BaSO_(4) = [Ba^(2+)][SO_(4)^(2-)]=(8xx10^(-4))(8xx10^(-4))=64xx10^(-8)` <br/> As `H_(2)SO_(4)` <a href="https://interviewquestions.tuteehub.com/tag/ionizes-7690815" style="font-weight:bold;" target="_blank" title="Click to know more about IONIZES">IONIZES</a> completely as `H_(2)SO_(4) rarr 2H^(+) + SO_(4)^(2-),[SO_(4)^(2-)]` producedfrom 0.01 M `H_(2)SO_(2-)]` produced from 0.01 M `H_(2)SO_(4)=0.01 M `. Thus, if S is the solubility of `BaSO_(4) ` in `H_(2)SO_(4) ` , then <br/> `K_(ap) = [Ba^(2+)][SO_(4)^(2-)] = S (S+ 0.01 ) = 64 xx 10^(-8) `(calculated above ) <br/>or `S^(2)+0.01 S - 64 xx 10^(-8) = 0 ` <br/> `:. S= (-0.01 <a href="https://interviewquestions.tuteehub.com/tag/pm-1156895" style="font-weight:bold;" target="_blank" title="Click to know more about PM">PM</a> sqrt((0.01)^(2)+4xx64xx10^(-8)))/(2) = (-0.01 pm sqrt(10^(-4)+256xx10^(-8)))/(2)` <br/> `=(-0.01 pm sqrt(10^(-4)(1+256xx10^(-2))))/(2) = (-0.01 pm 10^(-2) sqrt(1+256xx10^(-2)))/(2)` <br/> `=(-0.01 pm 10^(-2)sqrt(1.256))/(2) = (-10^(-2)+1.12xx10^(-2))/(2)` <br/> `=((-1+1.12)xx10^(-2))/(2)` <br/> `=((-1+1.12)xx10^(-2))/(2) = (0.12)/(2) xx 10^(-2) = 6 xx 10^(-4) ` mol `dm^(-3)`.</body></html>


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