Question

In a titration, 10.00 mL of .1000M butanoic acid, CH3CH2CH2COOH (Ka = 1.54 * 10^-5), is titrated with .1000M NaOH (aq). a. Calculate the pH of the system when 10.00 mL of NaOH (aq) has been added. Use ICF/ICE tables to support your answer. b. Sketch the general shape of the titration curve for this system. Be sure to label the axes, any buffer regions, the point at which pH = pKa and the equivalence point. You do not need to provide specific pH values on the curve.

Answer 1

Initial moles of acid = Molarity*Volume = 0.1*0.01 = 0.001

Moles of base NaOH added = 0.1*0.01 = 0.001

So this is the equivalence point.

At this point, conc of CH₃CH₂CH₂COO^- Na⁺ = C = 0.1/2 = 0.05 M

At equivalence point, we have:

[OH-] = (K_b*C)^0.5 = ((10^-14/K_a)*C)^0.5 = ((10^-14/(1.54*10^-5))*0.05)^0.5 = 5.69*10^-6 M

So,

pOH = 6-log(5.69) = 5.24

So,

pH = 14-5.24 = 8.76

The general titration curve is as shown below:

Titration Curve 14 12 10 excess base *- equivalence pt. r8 Buffer region 4 1/2 way pt. pH-pKa 0 0 12.5 25 37.5 50 62.5 75 87.5 mL of NaOH

Hope this helps !