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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 8 xx 10^(-4) "mol dm"^(-3). Calculate its solubility in 0.01 mol "dm"^(-3) of H_2SO_4.

Answer» <html><body><p></p>Solution :`BaSO_(4(g)) hArr Ba_((aq))^(2+) + SO_(4(aq))^(2-)` <br/> `K_(sp)` for `BaSO_4 = [Ba^(2+)] [ SO_4^(2-) ] =sxxs =S^2` <br/> But `S=<a href="https://interviewquestions.tuteehub.com/tag/8xx10-1931283" style="font-weight:bold;" target="_blank" title="Click to know more about 8XX10">8XX10</a>^(-4) "mol dm"^(-3)` <br/>`<a href="https://interviewquestions.tuteehub.com/tag/therefore-706901" style="font-weight:bold;" target="_blank" title="Click to know more about THEREFORE">THEREFORE</a> K_(sp)=(8xx10^(-4))^2 = 64xx10^(-8)` <br/> In the presence of 0.01 `MH_2 SO_4`, the expression for `K_(sp)` will be <br/> `K_(sp)=[Ba^(2+)][SO_4^(2-)]`<br/> `K_(sp)`=(s)(s+0.01)(0.01M `SO_4^(2-)` <a href="https://interviewquestions.tuteehub.com/tag/ions-1051295" style="font-weight:bold;" target="_blank" title="Click to know more about IONS">IONS</a> from 0.01 M `H_2SO_4` ) <br/>64 x 104 = s. (s+0.01) <br/> `s^2=0.01s-64xx10^(-8)=0` <br/> `s=(-0.01pmsqrt((0.01)^2+(4xx64xx10^(-8))))/2` <br/> `=(-0.01pm sqrt(10^(-4)+256xx10^(-8)))/2` <br/> `=(-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(10^(-4)(1+256 xx10^(-4))))/2` <br/> `=(-0.01 pm 10^(-2) sqrt(1+0.0256))/2` <br/> `=(10^(-2) (-1pm1.012719))/2` <br/> `=5xx10^(-3) (-1+1.012719)` <br/> `=6.4xx10^(-5) "mol dm"^(-3)`</body></html>


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