Calculate the pH during the titration of 20.00 mL

Question

Calculate the pH during the titration of 20.00 mL of 0.1000 M CH₃CH₂COOH(aq) with 0.2000 M LiOH(aq) after 7 mL of the base have been added. K_a of propanoic acid = 1.3 x 10^-5.

Answer 1

Henderson-Hasselbalch: Usual assumptions about weak acid and strong base:(limiting reagent)

pH = pK_a + log([PA^-]/[PA])

Assume moles of PA^- (conj. of PA) produced = moles of OH^- reacted)

.2moles/liter*.007liters = .0014 moles of PA^-

The acid moles is the original moles minus the moles neutralized by OH^-

.1 moles/liter * ,02 liters - .0014 moles = .0006 moles of PA

Now the new molarities have to be calculated: because the total volume is now 20 + 7 ml.

[PA^-] = .0014 moles/ (.027 liters)

[PA] = .0006 moles/ (,027 liters) Note that this step was unnecessary in this case.

Plug into H-H equation (pK_a = -log(K_a))

For the full Ka approach check out section on weak acid/strong base titration in your text.