Question

A. When nickel (II) nitrate dissolves in water the nickel (II) cation gets hydrated by six water molecules. It is this hydrated ion that acts as an acid, as seen in the equilibrium reaction above. What is the pH of a 5 M solution of nickel (II) nitrate?

B. What is the pH of a 1.14 M solution of CH₃COOH that is also 0.65 M in gallium acetate, the salt of it's conjugate base? The K_a for CH₃COOH = 1.8 X 10^-5.

C. What is the pH of a 0.457 M solution of potassium fluoride? Ka for HF = 7.1 X 10^-4.

Answer 1

A) Hydration of nickel (II) nitrate is given by the equation:

Ni(NO3)2 +6H2O --->[Ni(H2O)6](NO3)2

Acidic [Ni(H2O)6]²⁺ ion hydrolyses,pka=4.5

ka=10^-4.5=3.162*10^-5

[Ni(H2O)6]²⁺ + H2O <---> [Ni(H2O)5 OH^-]²⁺+H3O+

ka(nickel hexahydrate)=3.162*10^-5 =[Ni(H2O)5 OH^-]²⁺[H3O+]/[Ni(H2O)6]²⁺

^{ICE table,}

	[[Ni(H2O)6]2+	[Ni(H2O)5 OH-]2	[H3O+]
initial	5M	0	0
change	-x	+x	+x
equilibrium	5-x	x	x

ka=3.162*10^-5 =x^2/(5-x) .

5>>>x ,for very small ka value

3.162*10^-5 =x^2/(5)

0.0126M=x=[H3O+]

pH=-log [H3O+]=-log (0.0126)=1.9

pH=1.9