Question

A volume of 500.0 mL of 0.140 M NaOH is added to 585 mL of 0.250 M weak acid (K, = 1.73 x 10-). What is the pH of the resulting buffer? HA(aq) + OH(aq) — H, 0(1) + A- (aq) pH =

Answer 1

mmoles of NaOH = molarity ×volume

= 0.140×500 = 70

mmoles of weak acid (HA) = 0.250×585 =146.25

reaction is

HA + OH^- $\to$ H₂O +A^-

^{now BCA table is}

	HA	OH-	A-
before ( mmoles)	146.25	70	0
change (mmoles)	-70	-70	+70
after (mmoles)	146.25 - 70 = 76.25	0	70

given, Ka = 1.73×10^-5 , then pKa = - logKa = -log(1.76×10^-5 )

= 4.76

Solution contains excess weak acid and its conjugate base

using Henderson -Hasselbalch equation for this buffer

pH = pKa + log(mmoles of A^-/mmoles of HA)

or, pH = 4.76 + log (70/76.25)

or, pH = 4.76 - 0.037

or, pH = 4.723