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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. ) |
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Answer» 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 |
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