Determine the minimum concentration of KOH required for precipitation to begin i

Question

Determine the minimum concentration of KOH required for precipitation to begin in each case.
1) 1.7x10^-2 M CaCl2 ( ksp for Ca(OH)2: 4.68x10^-6)
2) 2.7x10^-3 M Fe(NO3)2 ( ksp for Fe(OH)2: 4.87x10^-17)
3) 1.9x10^-3 M MgBr2 ( ksp for Mg(OH)2: 2.06x10^-13)
Please explain the method for solving: I am currently trying ksp=[cation][anion] Determine the minimum concentration of KOH required for precipitation to begin in each case.
1) 1.7x10^-2 M CaCl2 ( ksp for Ca(OH)2: 4.68x10^-6)
2) 2.7x10^-3 M Fe(NO3)2 ( ksp for Fe(OH)2: 4.87x10^-17)
3) 1.9x10^-3 M MgBr2 ( ksp for Mg(OH)2: 2.06x10^-13)
Please explain the method for solving: I am currently trying ksp=[cation][anion]
1) 1.7x10^-2 M CaCl2 ( ksp for Ca(OH)2: 4.68x10^-6)
2) 2.7x10^-3 M Fe(NO3)2 ( ksp for Fe(OH)2: 4.87x10^-17)
3) 1.9x10^-3 M MgBr2 ( ksp for Mg(OH)2: 2.06x10^-13)
Please explain the method for solving: I am currently trying ksp=[cation][anion]

Explanation / Answer

a) CaCl2 + 2KOH ----> Ca(OH)2 + 2KCl

Ca(OH)2 <=> Ca2+ + 2OH-

ksp = [Ca,Fe,Mg2+][OH-]2

ksp = [1.7x10-2] [x2]
4.68x10-6 = [1.7x10-2][x2]

x= [KOH] = 1.6591x10-4

b) Fe(NO3)2 + 2KOH ----> Fe(OH)2 + 2KNO3

Fe(OH)2 <=> Fe2++ 2OH-

4.87x10-17 = [2.7x10-3][x2]

x = 1.3430x10-10

c) MgBr2 + KOH ----> Mg(OH)2 + 2KBr

Mg(OH)2 <=> Mg2+ + 2OH-

2.06x10-13 = [1.9x10-3][x2]

x = 1.0412x10-8

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