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Answer 1 · 2017-04-30T22:56:34

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

Molarity of NaOH

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

$HA + NaOH \to NaA + H_{2} O$ $"HA + NaOH" → "NaA" + "H"_2"O"$

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 $p K_{a}$ $pK_text(a)$

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

$HA + H_{2} O ⇌ H_{3} O^{+} + A^{-}$ $"HA "+ "H"_2"O" ⇌ "H"_3"O"^"+" + "A"^"-"$

$K_{a} = \frac{[H_{3} O^{+}] [A^{-}]}{[HA]}$ $K_text(a) = (["H"_3"O"^"+"]["A"^"-"])/(["HA"])$

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

$[A^{-}] = [HA]$ $["A"^"-"] = ["HA"]$

∴ $K_{a} = [H_{3} O^{+}]$ $K_text(a) = ["H"_3"O"^"+"]$

Taking the negative logarithm of both sides, we get

$p K_{a} = pH$ $pK_text(a) = "pH"$

At half-neutralization (10.0 mL),

$pH \approx 4.7$ $"pH" ≈ 4.7$ , so $p K_{a} \approx 4.7$ $pK_text(a) ≈ 4.7$