1.

Upto what pH must a solution containing a precipitate of Cr(OH)_(3) be adjusted so that all of precipitate dissolves

Answer»

Upto 4.4
Upto 4.1
Upto 4.2
Upto 4.0

Solution :`K_(sp) = [Cr^(2+)] [ OH^(-)]^(3)`
`[OH]^(-3) = K_(sp//Cr^(3+)) = (6 XX 10^(31))/(1 xx 10^(1)) = 6 xx 10^(-30)`
`[OH^(-)] = 1.8 xx 10^(-10)`
`pOH = (log1.8 + log10^(10)) = 10 + 0.25 + 1 = 11.25`
`pH = 14-11.25 = 2.27`


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