Question

A student peforms a titration, titrating 25.00 mL of a weak monoprotic acid, HA, with a 1.22 M solution of NaOH. They collect data, plot a titration curve and determine the values given in the below table.

	ml NaOH added	pH
Half-way Point	18.34	4.06
Equivalence point	36.68	8.84

How many moles of NaOH have been added at the equivalence point?
mol
What is the total volume of the solution at the equivalence point?
mL


During the titration the following reaction occurs

HA + NaOH ⇒ NaA + H₂O

What is the concentration of A^- (the conjugate base of the weak acid) at the equivalence point?
M
Use [A^-] and the pH at the equivalence point to estimate K_a of the weak acid (this is the hard way of estimating K_a)
  
Use the pH at the half-way point to estimate K_a of the weak acid (this is the easier way)

Answer 1

(1) the number of moles if NaOH added = molarity × volume (in L)

= 1.22mol/L ×(36.68×10^-3 L ) = 4.47496×10^-2mol

= 4.47×10^-2 mol

(2) Volume of solution at equivalent point = 25.00ml +36.68ml

= 61.68ml. (Answer)

(3)

[A^-] = number of moles/ volume in L = 4.47496×10^-2mol/(61.68×10^-3 L) = 0.725512 mol/L = 0.726 M. (answer)

(4)

pH = 7 + 1/2 × (pKa + log c)

8.84 = 7 + 1/2 ( pKa + log 0.726)

pKa = 3.68 - log 0.726 = 3.819355

pKa = 3.82

Ka = 10^-3.82 = 1.52×10^-4 (answer)

(5)

At half equivalence point , concentration of A^- and HA are same .

pH = pKa + log[A^-]/[HA]

pH = pKa = 4.06

Ka = 10^-4.06 = 8.71×10^-5. (Answer)