# Question cba6e

Apr 30, 2017

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

#### Explanation:

Molarity of NaOH

The equation for the reaction with $\text{NaOH}$ is

$\text{HA + NaOH" → "NaA" + "H"_2"O}$

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 $p {K}_{\textrm{a}}$

As long as we have some $\text{HA}$ present, we have the equilibrium

$\text{HA "+ "H"_2"O" ⇌ "H"_3"O"^"+" + "A"^"-}$

${K}_{\textrm{a}} = \left(\left[\text{H"_3"O"^"+"]["A"^"-"])/(["HA}\right]\right)$

At the mid-point of the titration (i.e. at 10.0 mL $\text{NaOH}$),

$\left[\text{A"^"-"] = ["HA}\right]$

${K}_{\textrm{a}} = \left[\text{H"_3"O"^"+}\right]$

Taking the negative logarithm of both sides, we get

$p {K}_{\textrm{a}} = \text{pH}$

At half-neutralization (10.0 mL),

"pH" ≈ 4.7, so pK_text(a) ≈ 4.7#