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Answer 1 · 2017-04-30T22:56:34

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

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 $pK_text(a) ≈ 4.7$