Answer:
CHEM 123 Solution to Ques 16.65 from Wiley #12
16.65 A student added 100 mL of 0.10 M NaOH to 250 mL of a buffer that contained 0.15 M HC2H3O2 and 0.25 M C2H3O2−. What is the concentration of HC2H3O2 after the addition of the strong base.
Begin by writing the equation for the buffer (acid dissociation of the weak acid)
HC2H3O2 + H2O H3O + + C2H3O2- Ka = 1.8 × 10-5
Initial amounts in the solution are:
# mol C2H3O2- = (0.25 mol/L)(0.25 L) = 0.063 mol C2H3O2-
# mol HC2H3O2 = (0.15 mol/L)(0.25 L) = 0.038 mol HC2H3O2
The added base (0.100 L)(0.10 mol/L) = 0.010 mol NaOH) will react with the acetic acid present in the buffer solution: NaOH + HC2H3O2 ⎯→ H2O + NaC2H3O2
Assume the added base reacts completely (since it's a strong base with a weak acid). For each mole of base added, one mole of HC2H3O2 is converted to C2H3O2-. Since 0.010 mol of base is added, 0.010 mol of HC2H3O2 will be depleted from the initial amount (0.038 mol) and 0.010 mol C2H3O2− ADDITIONAL produced in this reaction (added to the initial amount of 0.063 mol C2H3O2−):
# mol HC2H3O2final = (0.038 - 0.010) mol = 0.028 mol
# mol C2H3O2-final = (0.063 + 0.010) mol = 0.073 mol
The final volume of solution is 250 mL + 100 mL = 350 mL.
AFTER the reaction with NaOH, the concentrations become…
[HC2H3O2] = 0.028 mol/0.350L = 0.080 M HC2H3O2
[C2H3O2−] = 0.073 mol/0.350L = 0.21 M C2H3O2−
In Wiley #12, the question asked only what the concentration of acetic acid becomes, so this would be the answer: 0.080 M. (The optional given you in Wiley was 0.079 M.)
In the actual question in the book, it went further and asked you what is the CHANGE in concentration of HC2H3O2 and C2H3O2−, so you would simply go one further step:
Initial conc of HC2H3O2 was 0.15M
Change in conc = 0.15−0.080 = 0.0.07 M decrease in [HC2H3O2]
Initial conc of C2H3O2− was 0.25M
Change in conc = 0.25M − 0.21 Μ = 0.04 Μ decrease in [C2H3O2−]