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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»

` 1.5 xx 10^(-9) M`
`1.5 xx10^(-8) M`
` 2.25 xx 10^(-16)M`
` 2.25 xx 10^(-18) M`

Solution :` BaSO_4 hArr Ba^(+2)+SO_4^(-2) `
` 0.1 M BaCl_2 =0.1 M Ba^(+2) `
` [SO_4^(-2)]= S, [Ba^(+2) ]= (0.1 +S) ~~ 0.1M`
` K_(sp)=[Ba^(+2) ] [SO_4^(-2) ], 1.5 xx 10 ^(-9) =(0.1 ) (S) `
`RARR S = 1.5 xx 1 0^(-8) `


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