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.
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-] = (Kb*C)0.5 = ((10-14/Ka)*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:
Hope this helps !
In a titration, 10.00 mL of .1000M butanoic acid, CH3CH2CH2COOH (Ka = 1.54 * 10^-5), is...
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