Given : `Zn(OH)_2(s) hArr Zn(OH)_2(aq) , K_1=10^(-6)` `Zn(OH)_2(aq) hArr [Zn(OH)]^(+) +OH^(-) , K_2=10^(-7)` `[Zn(OH)]^+(aq) hArr Zn^(+2)+OH^(-) , K_3
Given : `Zn(OH)_2(s) hArr Zn(OH)_2(aq) , K_1=10^(-6)`
`Zn(OH)_2(aq) hArr [Zn(OH)]^(+) +OH^(-) , K_2=10^(-7)`
`[Zn(OH)]^+(aq) hArr Zn^(+2)+OH^(-) , K_3=10^(-4)`
`Zn(OH)_2(aq)+OH^(-) hArr [Zn(OH)_3]^(-) , K_4=10^(3)`
`[Zn(OH)_3]^(-) (aq) + OH^(-) hArr [Zn(OH)_4]^(2-) , K_5=10`
Find out the negative of loganithm of the solubility of solid `Zn(OH)_2` at `25^@C` at pH=6.Consider `Zn(OH)_2` makes saturated solution at `25^@C`
(Write down in OMR sheet if your answer is 1.23 then 1 2 3 etc. )
1 Answers
Correct Answer - 1
Dissolved `[Zn(OH)_2]=[Zn^(+2)]_(aq)+[Zn(OH)^(+)]_(aq)+(Zn(OH)_2)_(aq)+[Zn(OH)_3^(-)]+[Zn(OH)_4]^(2-)`
Now, `[Zn(OH)_2]_(aq)=10^(-6)` M in saturated solution.
so `[Zn(OH)]^(+)+(10^(-6)xx10^(-7))/([OH^-])=10^(-13)/([OH^(-)])`
Similarly `[Zn^(+2)]=10^(-17)/([OH^(-)]^2),[Zn(OH)_3]^(-)=10^(-3) [OH^(-)]`
`[Zn(OH)_4^(2-)]=K_5[Zn(OH)_3^(-)][OH^-]=(10^(-2)M^(-1))[OH^(-)]^(2)`
Dissolved `Zn(OH)_2=10^(-17)/([OH^(-)]^(2))+10^(-13)([OH^(-)])+10^(-6)+10^(-3)[OH^(-)]+10^(-2)[OH^(-)]^2`
`=10^(-17)/10^(-19)+10^(-13)/10^(-8)+10^(-6)+10^(-3)xx10^(-8)+10^(-18)=10^(-1)+10^(-5)+10^(-6)+10^(-11)=10^(-1)`
-log `Zn(OH)_2(aq)`=1