Chapter 15 Source Answers and Examples
Chapter problems → · Original reading →
This chapter has no answer key. Its worked examples follow.
Worked examples and reviewed revisions
Example Problem 6.5.
Worked example
Example Problem 6.5. — AP edition reviewed replacement.
Reviewed hypothetical 25 °C fixed-volume concentration model: find analytical HCl concentration for pH 1.00 in (i) otherwise pure water and (ii) a solution with 10.0 M total acetate-family concentration and pKa=4.76. Assume ideal activities and complete HCl dissociation; the concentrated model is hypothetical, not a real-solution prediction.
Reviewed hypothetical model: use ideal concentrations, fixed final volume, total acetic-acid-family concentration C=10.0 M, pKₐ=4.76 and target [H⁺]=0.100 M at 25 °C. This concentrated solution is not a realistic ideal buffer; the exercise diagnoses the original charge-balance error.
Mass balance gives [HA]+[A⁻]=10.0 M. Kₐ=[H⁺][A⁻]/[HA], so [A⁻]=CKₐ/([H⁺]+Kₐ).
Kₐ=10⁻⁴·⁷⁶=1.73780×10⁻⁵. Thus [A⁻]=10.0Kₐ/(0.100+Kₐ)=0.001737499 M and [HA]=9.998262501 M.
For the hypothetical final solution, charge balance is [H⁺]=[A⁻]+[Cl⁻]+[OH⁻]. With Kw=10⁻¹⁴, [OH⁻]=10⁻¹³ M.
[Cl⁻]=0.100−0.001737499−10⁻¹³=0.098262501 M. With complete HCl dissociation, this is the analytical HCl concentration in the final fixed-volume model.
The original 0.1114 M result lost the initial acid contribution and does not satisfy charge balance. The ideal-model requirement is approximately 0.0983 M HCl, compared with 0.100 M for otherwise pure water at this target pH.
For otherwise pure water, [Cl⁻]=[H⁺]−[OH⁻]=0.100−10⁻¹³≈0.100 M. Weak acid alone is not a buffer containing appreciable acid and conjugate base.
Example 6.5.1
Worked example
Example 6.5.1 — AP edition reviewed replacement.
Reviewed hypothetical 25 °C final-volume concentration model: the added analytical NaOH and HCl amounts correspond to 5.00 M and 4.00 M in the same final volume. Treat activities as ideal. Find pH after neutralization.
H⁺+OH⁻→H₂O consumes 4.00 M equivalents of each. Approximately 1.00 M OH⁻ remains, with Na⁺/Cl⁻ spectators.
Equilibrium [H⁺] is positive: [H⁺]=Kw/[OH⁻]≈10⁻¹⁴ M. The stoichiometric value “no excess acid” is not zero equilibrium hydronium.
pOH≈−log₁₀(1.00)=0.00, so pH≈14.00 using pKw=14.00 at 25 °C. Nonideal concentrated solutions require activities.
Example Problem 7.2.1
Original example and steps
Example Problem 7.2.1:
You are titrating 750mL of an acetic acid solution with 10M Potassium Hydroxide. The titration curve is shown below. What is the concentration of acetic acid?
Step 1: Identify the equivalence point.
As we can see, the increases significantly after the addition of about 20mL of the . Therefore, the equivalence point occurs after 20mL of titrant addition.
Step 2: Using the equivalence equation, solve for the concentration of the Acetic Acid?
It doesn’t matter what units you use for volume, so long as you keep it consistent.
Example Problem 7.2.2
Worked example
Example Problem 7.2.2 — AP edition reviewed replacement.
Reviewed hypothetical 25 °C ideal stoichiometric task: neutralize 0.800 L of 15.0 M HCl using 4.00 M NaOH. Find titrant volume and ideal equivalence pH, assuming complete dissociation, 1:1 neutralization and no other acid/base species.
n(HCl)=15.0×0.800=12.0 mol. HCl+NaOH→NaCl+H₂O is 1:1, so 12.0 mol NaOH is required.
V(NaOH)=12.0/4.00=3.00 L=3000 mL. In M·mL units, 15.0×800=12000, not 1200; the retained 300 mL result is superseded.
At ideal strong-acid/strong-base equivalence at 25 °C, [H₃O⁺]=[OH⁻] and pH=7.00. Neutral pH depends on temperature and activities.