Will the calculated Molarity of NaOH be too high or too low if the following hap

Question

Will the calculated Molarity of NaOH be too high or too low if the following happen.

Assume the error was not corrected. For each case give an explanation for your

conclusion. In your discussion consider how the situation affects the calculation of

molarity of NaOH and moles NaOH, which is calculated from the mass of KHP. (see

(e) below for example.)

a) KHP is wet when you weigh it.

b)You add the weighed KHP to a flask containing a small amount of distilled

water.

c) The buret has water in it when you add the NaOH.

d) You exceed the equivalence point by one milliliter.

I understood a but both a & b but I'm stuck on c & d..please help. Also, in the experiment is says that its very important that all water be removed and that only the standard solution be present...why? (Basically answers c)

Explanation / Answer

it will affect the result.

You use the mass of KHP you weigh out to calculate the moles of KHP you weigh out (moles = mass / molar mass.

If the KHP is a little moist, then not all the sample is actually KHP, so when you use the mass to determine the moles of KHP you get a higher number of moles then you actually have present. (because some of that mass is not KHP but you have assumed it is)

So , you think you have a certain number of moles, but there are actually less present then you think there are.

When you do the titration, NaOH reacts in a 1:1 ratio with KHP. So you determine there are the same number of moles of NaOH is the volume you use as there were KHP in the mass you weighed out. Since you have overestimated moles of KHP, you will then overestimate the number of moles of NaOH.

So NaOH will appear higher concentration then it actually is.

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