Question

I have 0.500 Liters of water. To that 0.500 Liters, I add the following:

a. 0.100 moles HCl

b. 0.100 moles HOAc

c. 0.100 moles NH4Cl

d. 0.100 HF

e. 0.050 moles NaOH

f. 0.050 moles NaOAc

g. 0.050 moles Mg(OH)2

What is the pH of the resulting mixture of everything?

Please show individual steps, since this question is meant to be answered sequentially from a to g. Thanks!

Answer 1

I have 0.500 Liters of water. To that 0.500 Liters, I add the following:

a. 0.100 moles HCl

volume of water = 0.5 liter

moles of HCl = 0.100

moalrity = number of moles / volume in L

= 0.100/0.5

= 0.2 M

HCl is fully ionized thus it gives 0.2 H+

pH = - log [H+]

= -log 0.2

= 0.699

b. 0.100 moles HOAc

volume of water = 0.5 liter

moles of HOAc = 0.100

moalrity = number of moles / volume in L

= 0.100/0.5

= 0.2 M

CH3COOH is a weak acid. It does not fully dissociate in water to produce the same concentration of H+ ions.

CH3COOH , Ka = 1.8*10^-5

The formula for Ka:
Ka = [H+] * [CH3COO-] / [CH3COOH]
We have to calculate [H+] We make two very safe assumptions:
[H+] = [CH3COO-]
And because [CH3COOH >>>> [H+] we can use the molarity of the original acid for [CH3COOH]
Substitute:
1.8*10^-5 = [H+]² /0.2
[H+]² = 3.6*10^-6
[H+] = 1.89*10^-3

pH = -log [H+]
pH = -log (1.89*10^-3)
pH = 2.72

c. 0.100 moles NH4Cl

volume of water = 0.5 liter

moles of NH4Cl = 0.100

moalrity = number of moles / volume in L

= 0.100/0.5

= 0.2 M

NH4Cl = NH4 +   Cl-

Molarity . . . . .NH4+ + H2O <==> H3O+ + NH3
Initial . . . . . . . .0.2 . . . . . . . . . . . . . .0 . . . . . .0
Change . . . . . .-x . . . . . . . . . . . . . . .x . . . . . .x
At Equil. . . . .0.2-x . . . . . . . . . . . . . .x . . . . . .x

The Kb for NH3 = 1.8 x 10^-5. Ka for NH4+ ion = (Kw / Kb) = (1 x 10^-14 / 1.8 x 10^-5) = 5.6 x 10^-10

Ka = [H3O+][NH3] / [NH4+]

5.6 x 10^-10 = x^2 / (02 - x)

Due to small value of Ka, x will be small compared to 0.2 so we can delete it from the 0.2 - x term.

x^2 / 0.2 = 5.6 x 10^-10
x^2 = 1.12x 10^-10
x = 1.06x 10^-5

= [H3O+]

pH = -log[H3O+] = -log (1.06x 10^-5)

= 4.98

d. 0.100 HF

volume of water = 0.5 liter

moles of HF= 0.100

moalrity = number of moles / volume in L

= 0.100/0.5

= 0.2 M

First calculate the [H+] for the acid
Ka = [H+] [F-] / [HF]
We know that [H+] = [F-] , and because dissociation is small, we take [HF] = 0.20
7.2*10^-4 = [H+]² / 0.2
[H+]² = (1.44*10^-4)
[H+] = 0.012
pH = -log [H+]
pH = -log ( 0.012)
pH = 1.92