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Conductivity of saturated solution of BaSO_(4)" at 315 K is "3.648 xx 10^(-6)" ohm"^(-1)" cm"^(-1) and that of water is 1.25 xx10^(-6)" ohm"^(-1)" cm"^(-1). Ionic conductance of Ba^(2+) and SO_(4)^(2-)" are 110 and 136.6 ohm"^(-1)" cm"^(2)" mol"^(-1) respectively. Calculate the solubility of BaSO_(4) in g/L. |
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Answer» Solution :`Lambda_(m)^(@)(BaSO_(4))=Lambda_(m)^(@)BA^(2+)+Lambda_(m)^(@)SO_(4)^(2-)=110+136.6="246.6 OHM"^(-1)"cm"^(-1)` `K_(BaSO4)=K_(BaSO4)" (solution)"-K_("water")=3.648xx10^(-6)-1.25xx10^(-6)` `=2.398xx10^(-6)"S cm"^(-1)` `Lambda_(m)^(c )=(Kxx1000)/("Solubility")=(2.398xx10^(-6)xx1000)/(246.6)=9.72xx10^(-6)"mol/L"` `"Solubility "=9.72xx10^(-6)xx233=2.26xx10^(-3)g//L` |
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