Question

Given a 0.1 M NH4Cl solution: a) Which is the pH of this solution? b) If 100 mL of 0.1 M HCl solution are added to 100 mL of the original solution, which is the pH? c) And if 100 mL of 0.1 M NH3 are added to 100 mL of the original solution? Data: kb(NH3 ) = 1.8 10^-5 .

Answer 1

Ans :-

(a). 0.1 M NH₄Cl solution :-

NH₄Cl is the salt of weak base NH₄OH and strong acid HCl and therefore its pH can be calculated by using the formula :

pH = 1/2 [pK_w - logK_b - logC] ................(1)

Where, K_w = Ionic product of water whose value = 1.0 x 10^-14 , So pK_w = - logK_w = -log 1.0 x 10^-14 = 14.

K_b = Dissociation constant of base, So pK_b = -logK_b = - log 1.8 x 10^-5 = 4.74

C = Concentration of salt = 0.1 M, So logC = log 0.1 = -1

Put all these values in equation (1) :

pH = 1/2 [14 - 4.74 - (-1)]

pH = 1/2 [14 - 4.74 + 1]

pH = 1/2 [10.26]

pH = 5.13

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(b). Addition of 100 mL of 0.1 M HCl solution :-

Number of moles = Molarity x Volume of solution in L

Therefore,

No. of moles of NH₄Cl = 0.1 M x 0.1 L = 0.01 mol

Total volume = 200 mL = 0.2 L

Therefore, Concentration of NH₄Cl = 0.01mol/0.2 L = 0.05 M

Again,

pH of NH₄Cl = 1/2[14 - 4.74 - log0.05 M]

pH = 1/2[14-4.74+1.30]

pH = 5.28

and

[H⁺] = 10^-pH = 10^-5.28 = 5.25 x 10^-6 M

Moles of H⁺ in NH₄Cl = 5.25 x 10^-6 M x 0.1 L = 5.25 x 10^-7 mol

and

moles of HCl or H⁺ = 0.1 M x 0.1 L = 0.01 mol

moles of H⁺ in 200 mL of solution = 0.01

Because, moles of H⁺ in 100 mL of the solutions are very less as compare to moles of H⁺ in 100 mL of HCl. Therefore, consider the moles of H⁺ only.

Moles of H⁺ in overall solution = 0.01 mol

and the

concentration of H⁺ = [H⁺] = 0.01 mol/0.2 L = 0.05 M

pH = - log [H⁺]

pH = - log 0.05 m

pH = 1.30

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(c). Addition of 100 mL of 0.1 M NH₃ solution :-  

ICE table of NH₃ is :

..........................NH₃ ...............+.................H₂O <--------------------> NH₄⁺...................+..............OH^-

Initial..................0.1 M.......................................................................0.0 M.....................................0.0 M

Change...............-x.............................................................................+x...........................................+x

Equilibrium .......(0.1-x) M.....................................................................x M.........................................x M

Expression of K_b is :

K_b = [NH₄⁺].[OH^-] /[NH₃]

1.8 x 10^-5 = x²/(0.1-x)

x² + 1.8 x 10^-5 x - 1.8 x 10^-6 = 0

On solving

x = 0.00133

because, [OH^-] = 0.00133

So,

[H⁺] = 1.0 x 10^-14 / 0.00133 = 7.52 x 10^-12 M

Moles of H⁺ in NH₃ = 7.52 x 10^-12 M x 0.1 L = 7.52 x 10^-13 mol

and

Moles of H⁺ in NH₄Cl = 5.25 x 10^-6 M x 0.1 L = 5.25 x 10^-7 mol

Because moles of H⁺ in NH₃ are very less, therefore,

Total moles of H⁺ in solution = 5.25 x 10^-7 mol

and

[H⁺] = 5.25 x 10^-7 mol / 0.2 L = 2.62 x 10^-6 M

pH = - log [H⁺]

pH = - log 2.62 x 10^-6 M

pH = 5.58