Please write the chemical equation of this equilibrium.

Question

An unknown chemical (HB) was analysed. It is known that this chemical is a weak base. 1.94g of pure HB was completely dissolved in 32.00mL water. As it is titrated against 0.753M hydrobromic acid (HBr), 32.60mL was used to reach the equivalence point.

(c) As HB was dissolved in water, it completely dissolves and undergoes partial protonation when accepting protons from water to form H₂B⁺ and OH^- . Please write the chemical equation of this equilibrium.

(d) Given that the equilibrium of base dissociation of HB (Kb) is 1.7x10-9 M, please evaluate the percentage of ionization of HB if 2.61g of HB dissolves into pH 9.5 NaOH.

Answer 1

(c)  HB(aq) + H₂O(l) ⇌ H₂B⁺(aq) + OH⁻(aq)

That is the standard weak-base equilibrium: the base takes a proton from water, which is what leaves OH⁻ behind. Water is the proton donor here, so it must appear on the left.

First, get the molar mass from your titration data

At equivalence, moles of HBr equal moles of HB (1:1 — HB accepts one proton):

n = (0.03260 L)(0.753 M) = 0.02455 mol
M = 1.94 g / 0.02455 mol = 79.0 g/mol

Worth pausing on that number. 79.0 g/mol with K_b = 1.7 × 10⁻⁹ identifies the unknown as pyridine, C₅H₅N (M = 79.10, K_b = 1.7 × 10⁻⁹). Both values match, which is a satisfying confirmation that the titration was handled correctly — and it explains the "HB" notation, since pyridine accepts one proton at its ring nitrogen to become pyridinium.

(d) Percent ionisation in pH 9.5 NaOH

n(HB) = 2.61 g / 79.0 g/mol = 0.0330 mol

⚠️ The solution volume is not given in the part you posted. I will use 1.00 L; substitute yours at this step if it differs.

[HB]₀ = 0.0330 M, and the NaOH supplies [OH⁻] = 10^−4.5 = 3.16 × 10⁻⁵ M

K_b = [H₂B⁺][OH⁻] / [HB], with x = [H₂B⁺]:

1.7 × 10⁻⁹ = x(3.16 × 10⁻⁵ + x) / (0.0330 − x)

x ≈ 1.69 × 10⁻⁶ M

% ionisation = (1.69 × 10⁻⁶ / 0.0330) × 100 = 0.0051%

Do not skip the NaOH here — unlike some problems, it matters

Run the same base in pure water and you get x = √(K_b × 0.0330) = 7.49 × 10⁻⁶ M, or 0.023% ionised.

So the pH 9.5 background cuts the ionisation by roughly a factor of 4.4. The reason is the common ion effect: OH⁻ is a product of the equilibrium, and the NaOH already supplies 3.16 × 10⁻⁵ M of it — four times more than the pyridine itself would generate. By Le Châtelier that pushes the equilibrium left.

Compare this with a stronger base, where the added hydroxide would be swamped and could be ignored. Whether a common ion matters depends on how it compares to what the equilibrium produces on its own — always check that ratio rather than assuming either way.

And 0.005% ionised is a reminder of how weak K_b = 1.7 × 10⁻⁹ really is: over 99.99% of the pyridine remains as neutral molecules.