A solution is prepared by pipetting 50 mL of a 0.1625 0.0004 M HCI into a 250 ml

Question

A solution is prepared by pipetting 50 mL of a 0.1625 0.0004 M HCI into a 250 ml. volumetric flask using a transfer pipet, then filling the flask to the mark with water. Do error analysis for both a and b. Turn in a separate report for part a and for part b When you turn your work in, you should make sure that all your work is shown for full credit. Also, make a table of all integrals you did and their numerical values. And make a separate table that lists the error in each variable, calculated by you or not calculated by you. The reported errors in these tables should be to at least one extra significant digit. Last, separately report your final answers to each problem with the error in the standard format (as is done for the concentration of HCI solution shown above). Also, you will need to submit a hard copy of all your calculations as well as a separate table that lists the value with error for each variable that was used and/or calculated in this problem. Grades will be based on accuracy, neatness, completeness and ease of reading (your work and the reporting thereof should flow in a logical and easy-to-read manner). a. (35 points). Calculate the molar concentration of HCI and report with the error. Rank the contribution to the error in this dilution from the possible sources, the pipet, the flask and the error in the original concentration. b. (65 points) A piece of weigh paper was weighed out to be 0.3145 g on a scale that had an average error of plus or minus 0.4 mg. NaOH was then added to the weigh paper. The mass of the sample and weigh paper was 0.4285 g. The NaOH was then added to the solution of HCI prepared above, what would be the new molarity of H,O"? Report with error. (Assume the NaOH is pure and the volume of the solution stays the same.) Tolerances of Class A pipets and volumetric flasks. Volume (mL) Volumetric Flask (TC) Transfer Pipet (TD) Measuring Pipet (TD) 10 25 50 100 200 250 500 1000 2000 0.01 0.015 0.02 0.02 0.03 0.05 0.08 0.10 0.12 0.20 0.30 0.50 0.006 0.006 0.01 0.02 0.03 0.05 0.08 0.10 0.01 0.01 0.02 0.03 0.05 TC: to contain TD: to deliver

Explanation / Answer

a) Since solution of HCl is 0.1625 +- 0.0004 ,

if molarity is 1.625M and if it is 0.1625 + 0.0004 = 0.1629 ,

percentage error = 100-x

x = 0.1625/0.1629 x 100 = 99.7

percentage error = 100-99.7 = 0.3%

Most errors of come from weighing the solid or from measuring the liquid added in the dilution of HCl .

volumetric cylinder should be avoided as most of the time it has not appropriate volume ,infact a measuring flask should be taken.

b)Since piece of paper weigh 0.3145g +- 0.0004g (conversion from mg to g) ,

NaOH weigh - 0.4285g

if there is a error of 0.3145g +- 0.0004g then there is a posibility of 0.0004g more or less NaOH.

In that case there is a error of 0.09334%

since the HCl was added to new solution then total error is 0.39334%.

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