Question

What is the pH of a solution made by dissolving 2.81 grams of calcium fluoride in enough water to make 450 mL of solution? The Ka for HF is 6.8x10–4.

Write a chemical equation for the hydrolysis reaction that explains why an aqueous solution of CH3NH3Cl is acidic.

Calculate the pH of a 1.22×10-2 M solution of the decongestant ephedrine hydrochloride if the pKb of ephedrine (its conjugate base) is 3.86.

Answer 1

1)

Given that,

Mass of calcium fluoride = 2.81 g

Molar mass of caF2 = 78 g/mol

We know, moles = mass / molar mass

= 2.81 / 78

= 0.036 mol

Now, calculate the molarity of the calcium fluoride solution,

Molarity = moles /volume * 1000

= 0.036 / 450 * 1000

= 0.08 M

Let us consider a reaction,

CaF2 ---> Ca²⁺ +2F^-

As 1 mole of CaF2 gives 2 moles of F- ions.

So, 0.08 M solution of CaF2 will have = 0.08*2

= 0.16 M F- ions

The fluoride ions hydrolyze in water to form hydrofluoric acid as follows:

F^- +H₂O --->HF+OH^-

The hydrolysis constant for fluoride ion is

K_h = K_w / K_a

= 10^-14 / 6.8x10–4

= 1.4 x 10^-11

Now, calculate the concentration of OH^- ions released at equilibrium as follows:

K_h = [HF] [OH^-] / [F^-]

1.4 x 10^-11 = [OH^-]² / 0.16

[OH^-] =1.49 x 10^-6

Now, calculate the pOH of the solution ,

pOH = - log[OH^-]

= - log[1.49 x 10^-6]

= 5.82

Let's calculate the pH of the solution is,

pH = 14- pH

= 14 - 5.82

= 8.18

2)

In water CH3NH3Cl dissociates as follows:

CH3NH3Cl --- H2O--- > CH3NH3 + (aq) + Cl-(aq)

CH3NH3⁺ (aq) reacts with water to form H3O⁺, which make solution acidic.

CH3NH3 + (aq) + H2O (l) ---- > CH3NH2 (aq) + H3O (aq)