Can someone explain in detail how to get the answer? I\'m trying to understand a
ID: 703407 • Letter: C
Question
Can someone explain in detail how to get the answer? I'm trying to understand and I really hope you will be able to help me for all 3!
Thankyou!
1. The solubility product, Ksp, of Ni(OH) is 6x10 Calculate the molar solubility of Ni(OH)2 in pure water 2. Calculate the molar solubility ofNi(OH): in 1 .0x 10" mol L'i sodium hydroxide solution. (Assume, initially, that the solubility is much less than (OH ].) 3. The formation constant, Ks for the complex ion, [Ni(NHod2. is 1.0x108 and K/p for nickel(II) hydroxide, calculate the value of the equilibrium constant for: Ni(OH)2(s) + 6NH3 (aq)? [Ni(NH3)6]2+ (aq) + 2OH-(aq)Explanation / Answer
Ans 1
The balanced reaction
Ni(OH)2?Ni2+ + 2OH-
Equilibrium expression for the reaction
Ksp=[Ni2+] [OH-]2
Let the molar solubility of Ni(OH)2 = x
Molar solubility of [Ni2+] = x
Molar solubility of [OH-] = 2x
Ksp = x (2x)^2
6*10^-16 = 4x3
x = 5.313 * 10^-6
Molar solubility of Ni(OH)2 in water = 5.313 * 10^-6 M
Ans 2
Initial concentration of OH- = Concentration of NaOH
= 1*10^-4 mol/L
The balanced reaction
Ni(OH)2?Ni2+ + 2OH-
Equilibrium expression for the reaction
Ksp=[Ni2+] [OH-]2
Let the molar solubility of Ni(OH)2 = x
Molar solubility of [Ni2+] = x
Molar solubility of [OH-] = 2x
Ni(OH)2?Ni2+ + 2OH-
x x (10^-4 + 2x)
Ksp = x (10^-4 + 2x)^2
Initially solubility is less than [OH-]
6*10^-16 = 10^-8x
x = 6 * 10^-8
Molar solubility of Ni(OH)2 in NaOH = 6 * 10^-8 M
Ans 3
Ni(OH)2?Ni2+ + 2OH- Ksp = 6*10^-16
Ni2+ + 6NH3 ? [Ni(NH3)6]2+ Kf = 1*10^8
Overall reaction by adding the above reactions
Ni(OH)2 + 6NH3 ? [Ni(NH3)6]2+ + 2OH-
Equilibrium constant
Keq = Ksp x Kf = 6*10^-16 x 1*10^8
= 6*10^8