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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 PbSO_(4), in water is 0.303 g/l at 25^(@)C, its solubility product at that temperature is |
Answer» <html><body><p>` 10^(-4)M^(2) ` <br/>` 9.18 xx 10^(-4) M` <br/>` 10^(-6)M^(2)` <br/>` 9.18 xx 10^(-8) M^(2)` </p>Solution :` S= 0.303 g // lit ` <br/> ` <a href="https://interviewquestions.tuteehub.com/tag/mol-1100196" style="font-weight:bold;" target="_blank" title="Click to know more about MOL">MOL</a> . <a href="https://interviewquestions.tuteehub.com/tag/wt-1462289" style="font-weight:bold;" target="_blank" title="Click to know more about WT">WT</a> = 207 + 32 + 64 =303 ` <br/><a href="https://interviewquestions.tuteehub.com/tag/moles-1100459" style="font-weight:bold;" target="_blank" title="Click to know more about MOLES">MOLES</a> ` =( 0.303)/( 303)=10 ^(-3) `<br/>` S = 10 ^(-3)" <a href="https://interviewquestions.tuteehub.com/tag/mole-1100299" style="font-weight:bold;" target="_blank" title="Click to know more about MOLE">MOLE</a> " // "lit"` <br/> ` pbSO_4 hArr Punderset(S) b^(+2)+Sunderset( S) O ^(-2), S =10 ^(-3) ` <br/> ` K_(sp)= ( 10 ^(-3) M)^(2) =10 ^(-6)M^(2) `</body></html> | |