Question #cba6e

1 Answer
Apr 30, 2017

Answer:

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#