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A solution which remains in equilibrium with undissolved solute is said to be saturated. The concentration of a saturated solution at a given temperature is called solubility. The product of concentration of ions in a saturated solution of an electrolyte at a given temperature, is called solubility product (K_(sp)). For the electrolyte, A_(x),B_(y),:A_(x),B_(y(s)) rarr xA^(y+)+ y^(Bx-), with solubility S, the solubility product (K_(sp)) =x^(x)xxy^(y) xx s^(x+y). While calculating the solubility of a sparingly soluble salt in the presence of some strong electrolyte containing a common ion, the common ion concentration is practically equal to that of strong electrolyte. If in a solution, the ionic product of an clectrolyte exceeds its K_(sp), value at a particular temperature, then precipitation occurs. The solubility of BaSO_(4), in 0.1 M BaCl_(2), solution is (K_(sp), of BaSO_(4), = 1.5 xx 10^(-9)) |
Answer» <html><body><p>` <a href="https://interviewquestions.tuteehub.com/tag/1-256655" style="font-weight:bold;" target="_blank" title="Click to know more about 1">1</a>.5 xx 10^(-9) M` <br/>`1.5 xx10^(-8) M` <br/>` 2.25 xx 10^(-<a href="https://interviewquestions.tuteehub.com/tag/16-276476" style="font-weight:bold;" target="_blank" title="Click to know more about 16">16</a>)M` <br/>` 2.25 xx 10^(-18) M` </p>Solution :` BaSO_4 hArr Ba^(+2)+SO_4^(-2) ` <br/> ` <a href="https://interviewquestions.tuteehub.com/tag/0-251616" style="font-weight:bold;" target="_blank" title="Click to know more about 0">0</a>.1 M BaCl_2 =0.1 M Ba^(+2) ` <br/>` [SO_4^(-2)]= S, [Ba^(+2) ]= (0.1 +S) ~~ 0.1M` <br/> ` K_(sp)=[Ba^(+2) ] [SO_4^(-2) ], 1.5 xx 10 ^(-9) =(0.1 ) (S) ` <br/> `<a href="https://interviewquestions.tuteehub.com/tag/rarr-1175461" style="font-weight:bold;" target="_blank" title="Click to know more about RARR">RARR</a> S = 1.5 xx 1 0^(-8) `</body></html> | |