ost-Lab Questions I. Using the average molarity of NaOH calculate % error in you

Question

ost-Lab Questions I. Using the average molarity of NaOH calculate % error in your determination of the 2. If your experimental molar concentration of NaOH were not close to the accepted value molarity of NaOH if the true value of the molar concentration were 0.1 021 M what are two likely reasons for the discrepancy. Do not simply say "human error." Explain clearly and specify whether the cause of error would lead to a higher or lower experimental molar concentration of NaOH A buret is graduated with zero at the top and 50.00 mL at the bottom. If the buret is filled with water and has an initial buret reading of 1.08 mL and 35.28 mL is removed from the tip, what would the final buret reading be? 3.

Explanation / Answer

1/ %error in molarity of NaOH = (True value - experimetal value) x 100/True value

%error is always reported as +ve value

True value = 0.1021 M

So knowing the experimental value, we can easily calculate %error.

[Experimental value not mentioned above]

2. If the %error is higher, it could be due to incorrect measurement of the amount of NaOH initially taken to make the solution. A higher amount of NaOH taken would result into a higher molar concentration of the resulting NaOH solution.

3. For the calibrated buret

the final buret reading would be = 35.28 - 1.08 = 34.20 ml

So we have added 34.20 ml of titrant to the solution from the buret.

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