1.

The conductivity of a saturated solution of BaSO_(4) at 298 K , is found to be 3.758 xx 10^(-6) ohm^(-1) cm^(-1) and that of water used is 1.36 xx 10^(-6) ohm^(-1) cm^(-1) . Molar ionic conduntances at infinite dilution for Ba^(2+) and SO_(4)^(2-) ions are 110 and 136.6 ohm^(-1) cm^(2) mol^-1 respectively . Calculate the solubility of BaSO_(4) at 295 K (Atomic masses : Ba = 137 , S = 32 , O = 16)

Answer»


SOLUTION :`kappa (BaSO_(4)) = kappa ` (solution ) - `kappa` (water)
`= 3.758 xx 10^(-6) - 1.36 xx 10^(-6)`
`= 2.398 xx 10^(-6) ohm^(-1) cm^(-1)`
`Lambda_(m)^(@) = 110 + 136.6 = 246.6 ohm^(-1) cm^(2) mol^(-1)`
Molar MASS = `137 + 32 + 4 xx 16 = 233`
Solubility = `(kappa xx 1000)/(Lambda_(m))`
`(2.398 xx 10^(-6) xx 1000)/(246.6)`
`= 9.724 xx 10^(-6) mol L^(-1)`
or `= 9.724 xx 10^(-6) xx 233 = 2.266 xx 10^(-3) G L^(-1)`


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