An unknown solution of PhCOOH of 15.0 mL is titrated against a solution of 0.057 mol L-1 KOH. 12.7 mL of the base solution is required to reach the equivalence point. Calculate the concentration of the PhCOOH solution.

Question

An unknown solution of PhCOOH of 15.0 mL is titrated against a solution of 0.057 mol L-1 KOH. 12.7 mL of the base solution is required to reach the equivalence point. Calculate the concentration of the PhCOOH solution.

BlackTom AI · Accepted Answer

We start by identifying the reaction: PhCOOH (acid) neutralizes KOH (strong base) in a 1:1 mole ratio because PhCOOH is monoprotic.
Option by option, we evaluate the plausibility based on moles:

Option A: 0.067 mol L-1
- This would imply moles of acid = concentration × volume = 0.067 mol/L × 0.0150 L = 0.001005 mol.
- At equivalence, the moles of base added must equal the moles of acid. The base volume at equivalence is 0.0127 L, so the required base concentration would need to satisfy 0.057 mol/L × 0.0127 L = 0.0007239 mol, which is not equal to 0.001005 mol. Therefore this option overestimates the acid concentration.

Option B: 0.057 mol L-1
- If the acid concentration were 0.057 M, the moles of acid in 0.0150 L would be 0.057 × 0.0150 = 0.000855 mol.
- The moles of base delivered at equivalence are 0.057 × 0.0127 = 0.0007239 mol. Since 0.000855 mol ≠ 0.0007239 mol, this would not reach equivalence with the given base volume; it would imply a different equivalence volume or concentration, making this option inconsistent with the data.

Option C: 0.21 mol L-1
- A concentration as high as 0.21 M would give moles of acid = 0.21 × 0.0150 = 0.00315 mol.
- The base provides only 0.0007239 mol at equivalence, so this would be far from matching the required neutralization; obviously not compatible with the titration data.

Option D: 0.048 mol L-1
- Calculating moles of acid: 0.048 × 0.0150 = 0.00072 mol (specifically 0.00072 mol for a rounded figure).
- This quantity closely matches the moles of base required to reach equivalence, which is 0.057 × 0.0127 = 0.0007239 mol. The close agreement between 0.00072 mol and 0.0007239 mol supports that the acid concentration is 0.048 M.

Overall, the key steps are recognizing the 1:1 stoichiometry and converting the given volumes and concentrations into moles to compare the acid and base amounts at equivalence. The only option that aligns with the exact stoichiometric balance given the data is 0.048 mol L-1, since it yields a mole amount of acid that matches the mole amount of base delivered at equivalence (within rounding).

CHEM 110 Lecture Quiz Week 10

An unknown solution of PhCOOH of 15.0 mL is titrated against a solution of 0.057 mol L-1 KOH. 12.7 mL of the base solution is required to reach the equivalence point. Calculate the concentration of the PhCOOH solution.

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