Question

In a titration of 42.91 mL of 0.3566 M ammonia with 0.3566 M aqueous nitric acid, what is the pH of the solution when 42.91 mL of the acid have been added?

Answer 1

Given:-

molarity of ammonia (NH₃) = 0.3566 M = 0.3566 mole / L

volume of ammonia (NH₃) used = 42.91 mL  = 42.91/ 1000 = 0.04291 L

molarity of nitric acid (HNO₃) = 0.3566 M   = 0.3566 mole / L

volume of nitric acid (HNO₃) used = 42.91 mL = 42.91/ 1000 = 0.04291 L

pH of resultant solution = ?

As we know that

molarity of compound (M) = no. of moles of compound (n) / volume of solution (L)

no. of moles of compound (n) = molarity of compound (M) $\times$ volume of solution (L)

Since we know that nitric acid (HNO₃) is a strong acid therefore

no. of moles of nitric acid (HNO₃) (n_HNO3) = molarity of nitric acid (HNO₃) (M) $\times$ volume of solution (L)

no. of moles of nitric acid (HNO₃) (n_HNO3) = 0.3566 mole / L  $\times$ 0.04291 L

no. of moles of nitric acid (HNO₃) (n_HNO3) = 0.015302 moles

similarly

no. of moles of ammonia (NH₃) (n_NH3) = molarity of ammonia (NH₃) (M) $\times$ volume of solution (L)

no. of moles of ammonia (NH₃) (n_NH3) =  0.3566 mole / L  $\times$ 0.04291 L

no. of moles of ammonia (NH₃) (n_NH3) = 0.015302 moles

Since we know that  when weak base such as ammonia (NH₃) being titrated with a strong acid such as nitric acid (HNO₃), the neutralization reaction occur and that will produce the conjugate acid (NH₄⁺) of the weak base (NH₃), which will then react with water to produce hydronium ions, H₃O⁺ and this hydronium ion determine the pH of the solution.

NH_3(aq) + HNO_3(aq) $\rightarrow$   NH₄NO_3(aq)

above reaction can be written as follows:-

NH_3(aq) + HNO_3(aq) $\rightarrow$ NH₄⁺_(aq) + NO₃^-_(aq)

0.015302 moles   0.015302 moles   0.015302 moles

So

No. of moles of NH₄⁺ (n_NH4⁺) = 0.015302 moles

Also know that the total volume of solution

Total volume of solution (V_total) = volume of ammonia (NH₃) used + volume of nitric acid (HNO₃) used

Total volume of solution (V_total) = 0.04291 L + 0.04291 L

Total volume of solution (V_total) = 0.08582 L

therefore

molar concentration of NH₄⁺ i.e [NH₄⁺] = No. of moles of NH₄⁺ (n_NH4⁺) /  Total volume of solution (V_total)

molar concentration of NH₄⁺ i.e [NH₄⁺] = 0.015302 moles /  0.08582 L

molar concentration of NH₄⁺ i.e [NH₄⁺] = 0.1783 mole /L = 0.1783 M

So we know that molar concentration of NH₄⁺ would determine the molar concentration of hydronium ions, H₃O⁺

NH₄⁺_(aq) + H₂O_(l) $\rightleftharpoons$ NH_3(aq)    + H₃O⁺_(aq)

Initial 0.1783 0 0

Change -x +x + x

Equiulibrium  0.1783 - x     +x + x

Now

acid dissociation constant of NH₄⁺ (K_a) = [NH₃][H₃O⁺] / [NH₄⁺]

acid dissociation constant of NH₄⁺ (K_a) = x $\times$ x  / (0.1783 - x )

acid dissociation constant of NH₄⁺ (K_a) = x² / (0.1783 - x )

since

0.1783 - x  $\simeq$  0.1783

so above equation becomes

acid dissociation constant of NH₄⁺ (K_a) = x² / 0.1783 --------------------------------(1)

Also we know that

NH_3(aq) + H₂O_(l) $\rightleftharpoons$   NH₄⁺_(aq) + OH^-_(aq)

base dissociation constant of NH₃ (K_b) = [NH₄⁺][OH^-] / [NH₃ ]

1.8 $\times$ 10^-5 = [NH₄⁺][OH^-] / [NH₃ ]

i.e

base dissociation constant of NH₃ (K_b) = 1.8 $\times$ 10^-5

since we know that

K_aK_b = K_w

K_a = K_w / K_b

therefore

acid dissociation constant of NH₄⁺ (K_a) = K_w / base dissociation constant of NH₃ (K_b)

acid dissociation constant of NH₄⁺ (K_a) = 10^-14 / 1.8 $\times$ 10^-5

acid dissociation constant of NH₄⁺ (K_a) = 5.6 $\times$ 10^-10 ----------------------------------(2)

form equation no. 1 and 2 we get

5.6 $\times$ 10^-10 = x² / 0.1783

x² = 0.1783 $\times$ 5.6 $\times$ 10^-10

x² = 0.99848 $\times$ 10^-10

x = $\sqrt{}$ 0.99848 $\times$ 10^-10

x =  0.999 $\times$ 10^-5

therefore

[H₃O⁺] = 0.999 $\times$ 10^-5

As we know that

pH = - log[H₃O⁺]

pH = - log[0.999 $\times$ 10^-5]

pH = - [ log(0.999) + ( - 5log(10) ]

pH = 5log(10) - log(0.999)

pH = 5  - (- 4.36)

pH = 5  + 4.36

pH = 9.36 (i.e answer)