Solubility products Ks(Fe(OH)2) = 4,9 x 10 ^ -17 Ks(Fe(OH)3) = 2,8 x 10 ^ -39 Ionic product of water Ke = 1,0 x 10 ^ -14 at 25 degrees
Solution
Molarity of Fe(+2) = Number of moles of FeSO4/Volume of solution in L = 0.100M
Ksp(Fe(OH)2) = [Fe(+2)][OH-]^2 = 4.9 * 10^(-17)
[OH-] = sqrt(4.9 * 10^(-16)) = 2.21 * 10^(-8)
pOH = 8 - log(2.21) = 7.65
pH + pOH = 14
pH = 14 - 7.65 = 6.35
Molarity of Fe(+3) = Number of moles of FeCl3/Volume of solution in L = 0.020M
Ksp(Fe(OH)3) = [Fe(+3)][OH-]^3 = 2.8 * 10^(-39)
[OH-] = (140* 10^(-39))^(1/3) = 5.1924 * 10^(-13)
pOH = 13 - log(5.1924) = 12.28
pH + pOH = 14
pH = 14 - 12.28= 1.72
Hence the Fe(OH)3 will precipitate first, since it rquire less amount of [OH-]
So in the pH range of 1.72-6.34, only one precipitate will be present
.
Solubility products Ks(Fe(OH)2) - 4-9 x 10 ^ -17 Ks(Fe(OH)3) - 2-8 x 1.docx
1. Solubility products Ks(Fe(OH)2) = 4,9 x 10 ^ -17 Ks(Fe(OH)3) = 2,8 x 10 ^ -39 Ionic product of
water Ke = 1,0 x 10 ^ -14 at 25 degrees
Solution
Molarity of Fe(+2) = Number of moles of FeSO4/Volume of solution in L = 0.100M
Ksp(Fe(OH)2) = [Fe(+2)][OH-]^2 = 4.9 * 10^(-17)
[OH-] = sqrt(4.9 * 10^(-16)) = 2.21 * 10^(-8)
pOH = 8 - log(2.21) = 7.65
pH + pOH = 14
pH = 14 - 7.65 = 6.35
Molarity of Fe(+3) = Number of moles of FeCl3/Volume of solution in L = 0.020M
Ksp(Fe(OH)3) = [Fe(+3)][OH-]^3 = 2.8 * 10^(-39)
[OH-] = (140* 10^(-39))^(1/3) = 5.1924 * 10^(-13)
pOH = 13 - log(5.1924) = 12.28
pH + pOH = 14
pH = 14 - 12.28= 1.72
Hence the Fe(OH)3 will precipitate first, since it rquire less amount of [OH-]
So in the pH range of 1.72-6.34, only one precipitate will be present
![Solubility products Ks(Fe(OH)2) = 4,9 x 10 ^ -17 Ks(Fe(OH)3) = 2,8 x 10 ^ -39 Ionic product of
water Ke = 1,0 x 10 ^ -14 at 25 degrees
Solution
Molarity of Fe(+2) = Number of moles of FeSO4/Volume of solution in L = 0.100M
Ksp(Fe(OH)2) = [Fe(+2)][OH-]^2 = 4.9 * 10^(-17)
[OH-] = sqrt(4.9 * 10^(-16)) = 2.21 * 10^(-8)
pOH = 8 - log(2.21) = 7.65
pH + pOH = 14
pH = 14 - 7.65 = 6.35
Molarity of Fe(+3) = Number of moles of FeCl3/Volume of solution in L = 0.020M
Ksp(Fe(OH)3) = [Fe(+3)][OH-]^3 = 2.8 * 10^(-39)
[OH-] = (140* 10^(-39))^(1/3) = 5.1924 * 10^(-13)
pOH = 13 - log(5.1924) = 12.28
pH + pOH = 14
pH = 14 - 12.28= 1.72
Hence the Fe(OH)3 will precipitate first, since it rquire less amount of [OH-]
So in the pH range of 1.72-6.34, only one precipitate will be present](https://image.slidesharecdn.com/solubilityproductsksfeoh2-4-9x10-17ksfeoh3-2-8x1-230201192448-037732bc/85/solubility-products-ksfeoh2-49-x-10-17-ksfeoh3-28-x-1docx-1-320.jpg)
