1.

Solubility product of Mg(OH)at ordinary temperature is 1.96 xx10^(-11). pH of a saturated soln. of Mg(OH)_2 will be

Answer»

`10.53 `
`8.47`
`6.94`
`3.47`

Solution :`MG(OH)_(2(s)) hArrMg^(2+) +2OH^(-)`
` K_(sp ) =([Mg]^(2+)[OH^(-)]^2)/( [Mg (OH)_2])= 1.96 xx 10^(-11)`
` x xx (2x)^2= 1.96 xx 10^(-11)` ( concentrationofsolid is UNITY )
`4x^3= 1.96 xx 10^(-11)`
` x=((1.96 xx 10^(-11))/(4) )^(1//3)`
` x= ( 4.9 xx 10^(-12))^(1//3)= 1.7 xx10^(-4)`
So ` OH^(-) ` concentration`= 2XX 1.7 xx 10^(-4)`
` i.e.,[OH^-] = 3.4 xx 10^(-4)`
Now `pOH=- log [OH^-]`
`=- log [3.4 xx 10^(-4)]=4- 0.531= 3.469 `
` therefore pH= 14 - 3.469= 10.531`


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