AAMC Test 4 #26

[riceguard]

Okay, so I seem to be clearly missing something because this doesn't make any sense to me

If a student did not remove all the moisture from the KHP before the titration with NaOH (aq), then the molarity determined for the NaOH (aq) would be:

The correct answer is A: too high because the actual number of moles of KHP titrated would be less than the number used in the calculations

But I went with D: unaffected because the weighted KHP was dissolved in water, making any moisture in the sample unimportant

I choose that because in the passage equation 2 is KHP(aq) + NaOH(aq) --> KNaP(aq) + H2O(l); KHP is aqueous, meaning it was dissolved in water, so what am I missing?

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[cafeverona]

I was tricked up on this too, and I think it's partly due to the wording of the question. From what I understand, they're talking about how the CALCULATED molarity of NaOH would differ from what actually happens in the experiment. Think of the calculation to find the molarity of NaOH. If your weight of KHP was too high, then your moles of KHP would be too high, and consequently, the calculated moles (and subsequently, molarity) of NaOH to titrate the solution would be too high compared to what you should see if you carried out the experiment correctly. Does that makes sense?