What is the concentration of an NH4Cl solution whose pH is 4.80? Kb for NH3 = 1.8 x 10^-5

• pOH= 14- 9.34= 4.66
• Here pKb= - log (1.75 x 10^-5)= 4.7570
• So pOH= pKb+ log (Salt/base)
• log(NH4Cl/NH4OH) =pOH- pKb
• = 4.66–4.757= - 0.097
• [NH4Cl/NH4OH]= 10^-0.097
• = 0.7998≈ 0.8
• Therefore the concentration of NH4Cl- should be 0.8- times the concentration of NH4OH.
• So number of moles of NH4Cl Tobe added = 0.5 x 0.8= 0.4- moles in 1- litre of the solution.
• = 53.5 x 0.4= 21.4- grams of NH4Cl is to be added to 1- litre SOLUTION containing 0.5- mole of NH4OH.

You can see from the pH that the ammonium chloride solution is acidic. This is due the presence of the ammonium ion which is the conjugate acid of ammonia, a weak base.

In water, the following equilibrium is established:

NH4+(aq) + H2O(l) = NH3(aq) + H3O+(aq)

The Ka for this system can be determined from the Kb of ammonia: Ka = Kw/Kb

So, Ka = 1.0x10^-14/1.8x10^-5 = 5.6x10^-10

The concentration of the NH4Cl solution will equal the concentration of the (NH4+)aq.

If we allow the initial concentration of NH4+ be represented by “y” and the change in concentrations that occur as the system reaches equilibrium be represented by “x”, then the equilibrium concentrations of the components can be expressed as follows:

[NH4+] = (y-x) [NH3] = x and [H3O+] = x

To simplify things, since the Ka for this system is very small we can assume that for the [NH4+] = (y-x) = y as the value of “x” will be very small.

Now because we know the pH of the system we can calculate the [H3O+] which will give us the value of “x”.

[H3O+] = 10^-pH = 1.58x10^-5 M = x

We can now substitute these values into the Ka equation and solve for “y”:

5.6x10^-10 = [NH3][H3O+]/[NH4+]

5.6x10^-10 = (x)(x)/y

5.6x10^-10 = (1.58x10^-5)^2/y

y =(1.58x10^-5)^2/5.6x10^-10 = 0.45M

So, the concentration of NH4Cl is 0.45 M

NH3 (ammonia) is a weak base.

NH3 + H2O <-> NH4+ + OH-

Initial 0.2M NH3, 0 M NH4+, 0 M OH-

Change x M NH3, x M NH4+, x M OH-

End 0.2 M NH3, x M NH4+, x M OH-

Kb = [NH4+][OH-]/[NH3] = 1.8 * 10^(-5)

x^2/(0.2) = 1.8 * 10^(-5)

x^2 = 3.6 * 10^(-6)

x = 1.9 * 10^(-3)

Thus [OH-} = 1.9 * 10^(-3) M

pOH = -log [OH} = - log (1.9 * 10^(-3)) = 2.7

pH = 14 - pOH = 14 - 2.7 = 11.3

[H+]^2 * Kb = Kwater * [NH4NO3]

[H+]^2 * 1.8×10^-5 = 10^-14 * 1.79

[H+] = 3.15×10^-5 M

pH = –log 3.15×10^-5

pH = 4.5

NH4Cl is the salt of a strong acid (HCl) and a weak base ( NH3) . The solution is therefore acidic - it Has pH < 7.00 . To calculate pH it is better to use the Ka value

Ka * Kb = 10^-14

Ka = 10^-14 / (1.8*10^-5)

Ka = 5.55*10^-10

The NH4+ ion hydrolyses: NH4+ + H2O → NH3 + H3O+

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

[NH3] = [H3O+] and [NH4+] = 0.4 M

we can write

5.55*10^-10 = [H3O+]² / 0.4

[H3O+]² = (5.55*10^-10) * 0.4

[H3O+]² = 2.22*10^-2

[H3O+] = 1.49*10^-5M

pH = - log [H3O+ ]

pH = - log 1.49/10^-5

pH = 4.83

[H+]^2 * Kb = Kwater * [NH4Cl]

[H+]^2 * 1.8 x 10^-5 = 10^-14 * 0.35

[H+]^2 = 1.94×10^-10

[H+] = 1.39×10^-5 M

pH = –log[H+]

pH = 5–log1.39

pH = 4.86

15g NH4Cl = 15/53.5 = 0.28 mol

15g NH3 = 15/17 = 0.88 mol

[OH–] = 0.88*1.8×10^-5 / 0.28 = 5.66×10^-5 M

pOH = –log 5.66×10^-5 = 4.25

pH = 14 – 4.25 = 9.75 (answer)

!!! No need to volume in this formula…

• pH of a salt of strong acid and a weak base is given by the formula
• pH= ½[pKw -pKb- logC]
• [Where Kw= Ionic product of water,
• Kb= Dissociation constantof NH3
• C= concentration of the salt]

1. here pKw= 14
2. pKb= — log(1.8 X 10^-5)= 4.7447
3. pH= 4.80
4. So 4.8 =½ [14–4.7447- logC]
5. logC =14–4.7447–9.6= -0.3447
6. So C = 10^ -0.3447= 0.4522 M

• A Very Simple method of calculation.