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Descriptive and laboratoryNational 2016

A student standardizes a solution of aqueous NaOH against a measured mass of solid potassium hydrogen phthalate. She then uses this NaOH solution to titrate a measured mass of an unknown monocarboxylic acid to its phenolphthalein endpoint to determine its molar mass. Which errors will lead to a value of the molar mass that is too high?

I. The potassium hydrogen phthalate is partially hydrated.

II. The NaOH solution is allowed to stand after being standardized and absorbs some carbon dioxide from the air.

18% of national qualifiers got this right

Explanation

draft

Step 1 – What raises or lowers the calculated molar mass?

Molar mass = mass of acid ÷ moles of acid, and moles of acid = (moles of NaOH used) = $M_{NaOH}\times V_{NaOH}$ .

So anything that makes the calculated $M_{NaOH}$ too high will make the calculated moles of acid too high, and therefore the reported molar mass too low (not too high). We need to check each error's effect on $M_{NaOH}$ .

Error I – Hydrated KHP

KHP's molar mass (used in the calculation) is the anhydrous value, $204.22\ \text{g/mol}$ . If the solid actually contains water, the true moles of KHP are:

$n_{true}=\frac{mass}{204.22+ (\text{extra mass from water})}$

which is less than the moles calculated with the anhydrous molar mass. Since $n_{NaOH}=n_{KHP}$ at the endpoint, the student calculates too many moles of NaOH for the volume used, so the computed $M_{NaOH}$ is too high.

When this

Correct trend for KHP hydration is right, but the student stops here without following the effect through to the unknown-acid titration, wrongly concluding the molar mass ends up too high.

Correctly identifies that CO2 absorption also inflates the calculated NaOH molarity, but again fails to carry the analysis through to the final molar-mass calculation, leading to the wrong high-molarity conclusion.

Assumes both errors independently raise the reported molar mass without checking that an inflated NaOH concentration actually overstates moles of acid, which drives the molar mass down, not up.

Correct: both errors make the calculated NaOH concentration too high. Using this inflated concentration in $n_{acid}=M_{NaOH}\times V_{NaOH}$ overstates the moles of acid, so molar mass = mass/moles is computed too low in both cases. Hence neither error yields a molar mass that is too high.

Concept · Acid-base standardization error propagation (KHP hydration and CO2 absorption effects on molar mass determination)