Question

Calculate the pH. Please show steps.
Nitric acid 0.00125 M	Acid
Potassium hydroxide 0.00133 M	Base
Ammonia 0.00120 M	Base
Hypochlorous acid 0.00128 M	Acid
Sodium hypochlorite 0.000125 M	Base
Ammonium chloride 0.00142 M	Salt
Carbonic acid 0.00144 M	Acid
Hydrochloric acid 4.55 M	Acid
Phosphoric acid 0.20 M	Acid

Answer 1

pH = - log 0.00125 = 2.9030(strong acid so 100% ionisation)

pOH = -log 0.00133 = 2.87614(strong base so 100% ionisation)

pH = 14 - pOH

so pH = 11.1238

Ammonia is a weak base with Kb = 1.75 × 10–5 and in this case cweak base >>> Kb, that is the equation to use is:

[OH–] =sqrt(K_b * Cweakbase)= SQRT(1.75 × 10–5 *0.00120) = 1.449 * 10^-4

pOH = -log[OH-] = -log(1.449*10^-4 ) = 3.83889

pH = 14 - pOH = 10.161

hypochlorus acid is a weak acid with Ka = 2.88 × 10–8 and in this case cweak acid >>> Ka, that is the equation to use is:

[H+] =sqrt(Ka * Cweakacid)= SQRT(2.88 × 10^–8 *0.00128) = 6.07 * 10^-6

pH = -log[H+] = -log( 6.07 * 10^-6 ) = 5.21669

Sodium hypochlorite is a weak base, a conjugate base of acetic acid, so:

k_b hypochlorite = 10^-14 / K_a( 2.88 × 10–8 ) = 3.47222222e-7

cweak base >>> Kb, that is the equation to use is:

[OH–] =sqrt(K_b * Cweakbase)= SQRT( 3.47222222e-7 * 0.000125) = 6.58807 * 10^-6

pOH = 5.18124

pH = 14 - pOH = 8.81875876249

NH4Cl is the salt of weak base ammonia (NH3) and the strong acid hydrochloric acid (HCl), therefore it is an acidic salt. That is, when it dissolves,
NH4Cl(aq) ---------> NH4+(aq) + Cl-(aq)
0.00142 M ................. 0.00142 M ...... 0.00142 M
Ammonium ion hydrolyzes, to produce acidic hydronium ion;

NH4+(aq) + H2O(l) <------> NH3(aq) + H3O+(aq)
0.00142 - x M ............................. x M .......... x M
Kh = [NH3][H3O+] / [NH4+]

If we find the value of Kh (hydrolysis constant) we can solve for x.

We know Kb.
NH3(aq) + H2O(l) <------> NH4+(aq) + OH-(aq)

Kb = [NH4+][OH-] / [NH3] = 1.8x10^-5

We also know the self-dissociation of water:
H2O(l) + H2O(l) <----> H3O+(aq) + OH-(aq)

Kw=[H3O+][OH-] = 1x10^-14

Kw / Kb = Kh

{ [H3O+][OH-] } / { [NH4+][OH-] / [NH3] } =
[H3O+][NH3] / [NH4+]

Kh = 1x10^-14 / 1.8x10^-5 = 5.55x10^-10

5.55x10^-10 = (x)(x) / (0.00142 - x)

Since x is very small compared to 0.00142, it can be neglected.

5.55x10^-10 = (x)(x) / (0.00142)

x^2 = 7.88x10^-13
x = 8.87749965e-7 M = [H3O+] or [H+]

pH = - log [H+] = - log (8.87749965e-7)
pH =6.05171

CO2(aq) + H2O(l) <==> H+ + HCO3- ........ Ka = 4.3x10^-7
0.00144M ........................... 0 .........0 ............... initial
-x ..................................... +x ....... +x ............. change
0.00144-x ........................... x .......... x ............... equilibrium

Ka = [H+] [HCO3-] / [CO2]
4.3x10^-7 = x² / (0.00144-x)
x = 2.48x10^-5

[H+] = 2.48 x10^-5
pH = -log[H+] = -log(2.48x10^-5) = 4.61 ........ to two significant digits


We shall ignore the ionization of HCO3- since it will make no measurable contribution to the hydrogen ion concentration.

Just in case you weren't sure about that....

HCO3- <==> H+ + CO3^3- ..... Ka = 5.6x10^-11
2.48x10^-5....2.48x10^-5..0 ...... initial
-x ................. +x ......... +x .............. change
2.48x10^-5-x 2.48x10^-5+x x ..... equilibrium

Ka = [H+] [CO3^2-] / [HCO3-]
5.6x10^-11 = (2.48x10^-5+x)(x) / (2.48 x10^-5-x)
x = 5.6x10^-11

the ionization of HCO3- makes no significant contribution to the hydrogen ion concentration.

H+ =4.55 M

pH = -log 4.55 =-0.66 (strong acid so 100% ionisation)

So pH = 1.462180