1.

K_(sp) of Ag_2CrO_4 is 1.1 times 10^-12. What is solubility of Ag_2CrO_4 in 0.1 M K_2CrO_4.

Answer»

Solution :`Ag_2CrO_4 leftrightarrow 2Ag^+ +CrO_4^(2-)`
x 2X x
x is the solubility of `Ag_2CrO_4` in 0.1 M `K_2CrO_4`
`K_2CrO_4 leftrightarrow 2K^+ + CrO_4^(2-)`
0.1 M 0.2 M 0.1 M
`[Ag^+]=2x`
`[CrO_4^(2-)]=(x+0.1) APPROX 0.1 therefore x lt lt 0.1`
`K_(sp)=[Ag^+]^2 [CrO_4^(2-)]`
`1.1 TIMES 10^-12=(2x)^2 (0.1)`
`1.1 times 10^-12=0.4x^2`
`x^2=(1.1 times10^-12)/0.4`
`x=sqrt((1.1 times10^-12)/0.4)`
`x=sqrt(2.75 times10^-12)`
`x=1.65 times 10^-6 M`


Discussion

No Comment Found

Related InterviewSolutions