Question & Answer: I'm having trouble figuring out titration for weak base/strong acid. The answers are in the…..

a) The reaction of NH₃ with water is as below:

NH₃ (aq) + H₂O (l) ——-> NH₄⁺ (aq) + OH^– (aq)

K_b = [NH₄⁺][OH^–]/[NH₃]

===> 1.8*10^-5 = (x).(x)/(0.1 – x)

Use small x approximation. Since K_b is small, we can assume x << 0.1 M and hence,

1.8*10^-5 = x²/0.1

===> x² = 1.8*10^-6

===> x = 1.3416*10^-3

[OH^–] = 1.3416*10^-3 M and hence pOH = -log [OH^–] = -log (1.3416*10^-3) = 2.8724

Therefore, pH = 14.00 – pOH = 14.00 – 2.8724 = 11.1276 ≈ 11.13 (ans).

b) Write down the reaction of NH₃ with HBr as below:

NH₃ (aq) + HBr (aq) ——-> NH₄⁺ (aq) + Br^– (aq)

As per the balanced stoichiometric equation,

1 mole NH₃ = 1 mole HBr = 1 mole NH₄⁺

Half way to the equivalence point, NH₃ is half neutralized and the number of moles of NH₃ retained at equilibrium is equal to the number of moles of NH₄⁺ formed.

Since the volume of the solution remains constant; [NH₄⁺]/[NH₃] = 1.

Use the Henderson-Hasslebach equation.

pOH = pK_b + log [NH₄⁺]/[NH₃] = -log (K_b) + log [NH₄⁺]/[NH₃] = -log (1.8*10^-5) + log (1) = 4.7447 + (0) = 4.75 (ans).

pH = 14.00 – pOH = 14.00 – 4.74 = 9.26 (ans).

c) At the equivalence point, NH₃ is completely neutralized and we have NH₄⁺ which is the conjugate acid of the weak base, NH₃.

Millimoles of NH₄⁺ formed = millimoles of NH₃ present initially = (30.00 mL)*(0.1 M) = 3 mmol.

Volume of HBr required for the titration = (millimoles of NH₃ present initially)/(concentration of HBr) = (3 mmol)/(0.2 M) = 15.00 mL.

Total volume of solution = (30.00 + 15.00) mL = 45.00 mL.

[NH₄⁺] = (millimoles of NH₄⁺)/(volume of solution) = (3 mmol)/(45.00 mL) = 0.06667 M.

The acid dissociation reaction is given as

NH₄⁺ (aq) <=====> NH₃ (aq) + H⁺ (aq)

Since H⁺ is formed, we shall use K_a.

K_a = [NH₃][H⁺]/[NH₄⁺] = (x).(x)/(0.06667 – x)

====> 5.5*10^-10 = x²/(0.06667 – x)

Use small x approximation. Since K_b is small, we can assume x << 0.06667 M and hence,

5.5*10^-10 = x²/0.06667

===> x² = 3.66685*10^-11

===> x = 6.05545*10^-6

[H⁺] = 6.05545*10^-6 M and pH = -log [H⁺] = -log (6.05545*10^-6) = 5.2178 ≈ 5.22 (ans).