Question

What is the pH of a 1.50 mol/L solution of ammonium chloride (NH4Cl)

2. What is the pH of a 1.50 mol/L solution of ammonium chloride (NH4Cl)?

kb = 1.8 10^-5

Answer 1

Answer-

NH₄Cl decomposes in water by following reaction -

NH₄Cl $\small \rightleftharpoons$ NH₃ + HCl

According to the stoichiometry of the reaction-

1 mol of NH₄Cl dissociates into 1 mol of NH₃ and 1 mol of HCl

Let the volume of the solution = 1L

Then the initial moles of NH₄Cl = 1.50

And let x moles of NH₄Cl dissociates at equilibrium to forms x mol of NH₃ and x mol of HCl

Now-

Time	NH4Cl moles	NH3 moles	HCl moles
t = 0	1.50 mol	0	0
At equilibrium	(1.50 - x) mol	x mol	x mol

K_b expression at equilibrium for this reaction is:

K_b = [NH₃][HCl] / [NH₃Cl]

The value of K_b is given as 1.8 x 10^-5

$\Rightarrow$

1.8 x 10^-5 = [NH₃][HCl] / [NH₃Cl]

1.8 x 10^-5 = x.x / (1.50 - x)

1.8 x 10^-5 = x² / (1.50 - x)

Since NH₄Cl is weak electrolyte here, the value of x will be very small

So we can say that, 1.50 - x $\small \approx$ 1.50

Now

1.8 x 10^-5 = x² / 1.50

1.8 x 1.50 x 10^-5 = x²

x² = 1.8 x 1.50 x 10^-5

x² = 2.7 x 10^-5

x² = 27 x 10^-6

x² = $\small \sqrt{27}$ x 10^-3

x = 5.196 x 10^-3

This means that 5.196 x 10^-3 mol of HCl is formed at equilibrium

And we know that HCl is a strong electrolyte and completely dissociates to H⁺ and Cl^- ions by following reaction-

HCl $\small \rightarrow$ H⁺ + Cl^-

Now according to the stoichiometry of the reaction-

1 mol of HCl forms $\small \rightarrow$ mol of H⁺

5.196 x 10^-3 of HCl forms $\small \rightarrow$ 5.196 x 10^-3 of H⁺

And the volume of the solution remains same (1L)

So the concentration of H⁺ ions = 5.196 x 10^-3 mol/L

Now these free H⁺ ions will contribute for the pH of the solution, and the pH value is given by the relation-

pH = -log[H⁺]

pH = -log[ 5.196 x 10^-3]

pH = 2.284

So the pH of this solution is 2.284