1.

What is the maximum pH of a 0.1 M Mg^(2+) solution from which Mg(OH)_2 will not be precipitated [K = 1.2 xx10^(-11)]?

Answer»

`11.02 `
`8.40`
`6.42 `
`9.04`

Solution :When ` K_(sp )=k_(sp)`precipitation doesnotoccur
` THEREFORE[Mg^(2+)][OH^(-)]^2 = 1.2xx 10^(-11)`
` [OH^-]= sqrt(( 1.2 XX 10^(-11))/(0.1 ))=1.09 xx 10^(-5)`
POH`=5- log = 4.96thereforepH = 14 -4.96= 9.04`


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