a. What is the [HF] in a 1.60 M solution of HF? When nickel (II) nitrate dissolv

Question

a. What is the [HF] in a 1.60 M solution of HF?

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?

c. 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 Ka for CH3COOH = 1.8 X 10-5.

d. What is the initial pH of a buffer that is 1.08 M CH3CH2CH2COOH and 0.99 M CH3CH2CH2COONa? The Ka for CH3CH2CH2COOH is 1.3 X 10-5.

e. What is the pH of a 1.06 M solution of HNO2 that is also 0.45 M in potassium nitrite, the salt of it's conjugate base? The Ka for HNO2 = 4.6 X 10-4.

Explanation / Answer

a) Given :[HF]o=initial concentration=1.60M

In solution the following equilibrium exists,

HF +H2O <--->F- +H3O+ (hydrolysis of HF)

ka=[F- ] [H3O+]/[HF].........(1)

ICE table

ka=[F- ] [H3O+]/[HF].........(1)

ka=7.2*10^-4=x^2/(1.60-x)

1.60>>>x ,as ka is small,very less acid dissociation takes place

7.2*10^-4=x^2/(1.60)

x=3.4*10^-2M=[H3O+]=[F-]

[HF]equilibrium=1.60M-x=1.60M-3.4*10^-2M=1.567M

[HF] at eqm=1.567M

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