A student titrates a weak monoprotic acid with 0.100 M NaOH
CHEMISTRY · Single correct · A Level · Chemistry · Acids and bases · Acids and bases
Problem
A student titrates a weak monoprotic acid with 0.100 M NaOH. Exactly half the moles of the acid have reacted when 25.0 mL of base has been added, and the measured pH is 4.38.
Which of the following best explains why the numerical value 4.38 is equal to the acid’s pKa?
(a) The ratio [HA]/[A−] equals 2 at this stage, making the Henderson–Hasselbalch logarithmic term equal to 1.
(b) At this volume the concentrations of the acid and its conjugate base are equal, so the logarithmic term in the Henderson–Hasselbalch equation is zero.
(c) The equivalence point has been reached, so the solution contains only the conjugate base, whose Kb equals the original acid’s Ka.
(d) Because [A−]/[HA]=0.50 at the midpoint, the term log(0.50) is often treated as negligible, so the Henderson–Hasselbalch equation gives pH≈pKa.