Question

Question #cba6e

Answer 1

B. `pK_text(a) = 4.7; "[NaOH] = 0.1 mol/L"`

Explanation:

Molarity of NaOH

The equation for the reaction with `"NaOH"` is

`"HA + NaOH" → "NaA" + "H"_2"O"`

It took 20.0 mL of NaOH to neutralize 20.0 mL of HA, so the two solutions must have had the same concentration, i.e. 0.10 mol/L.

Finding `pK_text(a)`

As long as we have some `"HA"` present, we have the equilibrium

`"HA "+ "H"_2"O" ⇌ "H"_3"O"^"+" + "A"^"-"`

`K_text(a) = (["H"_3"O"^"+"]["A"^"-"])/(["HA"])`

At the mid-point of the titration (i.e. at 10.0 mL `"NaOH"`),

`["A"^"-"] = ["HA"]`

∴ `K_text(a) = ["H"_3"O"^"+"]`

Taking the negative logarithm of both sides, we get

`pK_text(a) = "pH"`

At half-neutralization (10.0 mL),

`"pH" ≈ 4.7`, so `pK_text(a) ≈ 4.7`