Question

Consider the titration of 40.00 mL of 0.200 M NH₃ with 0.500 M HCl. Calculate pH
a) initially  
b) after addition of 5.00 mL of 0.500 M HCl
c) at the half-neutralization point (halfway to equilvalence point)
d) after the addition of 10.00 mL of 0.500 M HCl

e) at the equivalence point

Answer 1

a) Initial pH : Initially we have only 40.00 mL of 0.200 M NH₃ and no acid,

NH₃ is a weak base which forms NH4+ ions when dissolved in water.

Equlibrium Reaction can be written as NH3 + H2O NH₄⁺ + OH^-

Initial Conc,                                    0,200 M                0          0

Change                                   -x                     +x         +x

at Equlibrium                               (0,200 M -x)             x          x

The ionization constant of NH₃ (K_b) = 1.80×10⁻⁵ = [NH₄⁺][HO⁻] / [NH₃(aq)]

Thus 1.80×10⁻⁵ = [x][x] / (0.200-x)

assume 0.200 x

We have 1.80×10⁻⁵ = [x²] / 0.200 which means x= 1.9 x 10^-3 = [OH^-]

So Concentration of [OH^-] = 1.9 x 10^-3

As we know p^OH=−log₁₀[OH^-]=−log₁₀(1.9 x 10^-3)= 2.72

so pH = 14- p^OH = 14- 2.72 = 11.28 (Initial pH).

b) After addition of 5.00 mL of 0.500 M HCl :

NH3(weak base) will react with HCl (strong Acid) to produce ammonium cations, NH₄⁺, chloride anions, Cl⁻, and water. Reaction can be written as

NH_3(aq) +   H₃O⁺_(aq)   →   NH₄⁺ _(aq)+ H₂O_(l)

The hydronium cation H₃O⁺(aq) is produced by the dissociation of the HCl (strong Acid) in water.

So, We can see that for one mole of ammonia (consumed) will react with one mole of hydronium cations (Consumed) and generate one mole of ammonium cations (Produced) and the solution will be a "Ammonia/Ammonium Ion Buffer" (pKa = 9.24)

At the start, we have 40.00 mL of 0.200 M NH₃ , n(NH3) = 0.04 × 0.2 = 0.008 mols

We added, 5.00 mL of 0.500 M HCl, n(HCL) = 0.005 × 0.5 = 0.0025 mols

0.0025 mols of n(HCL) react will 0.0025 mols of n(NH3) to produce 0.0025 mols of salt i.e NH₄⁺ (aq).

So we have 0.0055 or 5.5 X 10^-3 mols of NH₃ left (base) and we have created 0.0025 mols or 2.5 X 10^-3 mols of NH₄+ (salt).

The total volume is now 40+5 = 45 ml, So [NH3] = 0.12 M and [NH₄⁺] = 0.05 M

pH= pKa+ log([NH3] / [NH4⁺])= 9.24 + log (0.12/0.05) = 9.24 + log (2.4) = 9.62.

c) at the half-neutralization point (halfway to equilvalence point) :

Initial moles of n[NH₃] = 40*0.2/1000 = 8 X 10^-3 mols

at half-neutralization point, half of the moles of n[NH₃] consumed = 4 X 10^-3 mols

so n[NH₃] left = 4 X 10^-3 mols and 4 X 10^-3 mols

At half equivalence Point, n[HCL] added = 4 X 10^-3 mols , Volume = 4 X 10^-3/ 0.5 = 8 ml

Total volume = 40+ 8 = 48 ml

n[NH₃] = n[NH₄₊] = 4 X 10^-3 mols

Concentration at half neutralization point [NH₃] = [NH_4⁺] = 4 X 10^-3 mols = 0.83 M

pH= pKa+ log([NH3] / [NH4⁺])= 9.24 + log (0.83/0.83) = 9.24 + log (1) = 9.24.

So at half neutralization point pH =pKa.

d) After addition of 10.00 mL of 0.500 M HCl : In the same manner,

At the start, we have 40.00 mL of 0.200 M NH₃ , n(NH3) = 0.04 × 0.2 = 0.008 mols

We added, 10.00 mL of 0.500 M HCl, n(HCL) = 0.010 × 0.5 = 0.005 mols

0.005 mols of n(HCL) react will 0.005 mols of n(NH3) to produce 0.005 mols of salt i.e NH₄⁺ (aq).

So we have 0.003 or 3 X 10^-3 mols of NH₃ left (base) and we have created 0.005 mols or 5 X 10^-3 mols of NH₄+ (salt).

The total volume is now 40+10 = 50 ml, So [NH3] = 0.06 M and [NH₄⁺] = 0.1 M

pH= pKa+ log([NH3] / [NH4⁺])= 9.24 + log (0.06/0.1) = 9.24 + log (0.6) = 9.01.

e) At equivalence Point :

Initial moles of n[NH₃] = 40*0.2/1000 = 8 X 10^-3 mols

At equivalence Point, n[HCL] = 8 X 10^-3 mols , Volume = 8 X 10^-3/ 0.5 = 16 ml

Total volume = 40+ 16 = 56 ml

n[NH3] =0 moles → completely consumed

n[H₃O⁺(aq)] =0 moles → completely consumed

n[NH⁺₄]=8 X 10^-3 moles → produced by the reaction

[NH⁺₄]= 0.14 M

Since it is a weak acid, we will use K_a to determine the H+ concentration.

Equlibrium Reaction can be written as NH₄⁺ + H2O NH₃ + H₃O⁺

Initial Conc,                                    0,14 M                0          0

Change                                   -x                     +x         +x

at Equlibrium                               (0,14 M -x)             x          x

The ionization constant of NH₄⁺ (Ka) = 5.7×10⁻¹⁰ = [NH₃][H₃O⁺] / [NH₄⁺]

Thus 5.7×10⁻¹⁰ = [x][x] / (0.14-x)

assume 0.14 x

We have 5.7×10⁻¹⁰= [x²] / 0.14 which means x²= 0.798 x 10^-10

x= 0.89x 10^-5 = 8.9 x 10^-6 = [H₃O⁺]

pH = - log[H₃O⁺] = - log (8.9 x 10^-6 ) = 5.06