How do I go about finding the data for numbers 9, 10, 11, 12, 13, and 14 in part

Question

How do I go about finding the data for numbers 9, 10, 11, 12, 13, and 14 in part IV if this lab. DO NOT WORRY ABOIT PART II. As the instructor told us not to do it. The data to figure out 9-14 is in numbers 4-6. Titration 1 Mass of KHP and weighing paper is 1.0527 grams. Mass of KHP is 0.527 grams. Saction Duter Data Sheet "N Delivering Solution from the Buret (1) final buret reading, mL (2) initial buret reading, mL (3) volume of H20 delivered, mL approved by approved by III. Preparing the KHP Sample Determination : ) (4) mass of weighing paper 1, 052 and KHP g (5) mass of weighing paper, g (6) mass of KHP g IV. Titrating the KHP (7) final buret reading, mL (8) initial buret reading, mL. (9) volume of NaOH used, mL (10) volume of NaOH used, L (11) number of moles of KHP titrated, mol (12) number of moles of NaOH SoPI required to to completely react with KHP, mol (13) molarity of the NaOH solution, mol L-1 (14) mean molarity of NaOH solution, mol L-1 123

Explanation / Answer

10)volume of NaOH used(L)

=(vol in ml)*(1L/1000ml)

11)mol of KHP titrated

KHP+NaOH-->NaKP+H2O(1:1 molar ratio of KHP and NaOH)

mol of KHP=mass of KHP/molar mass of KHP

=0.527g/204.22g/mol

=0.00258 mol

=0.5168g/204.22g/mol

=0.00253 mol

=0.5143g/204.22g/mol

=0.00252 mol

=0.00258 mol

0.00258mol/0.00778L

=0.332 mol/L

==0.00253mol/0.00792L

=0.319mol/L

0.00252 mol/0.00741L

=0.340 mol/L

9)volume of NaOH used(ml)( final burette-initial) 7.80-0.02=7.78ml 15.72-7.80=7.92 7.42-0.01=7.41

10)volume of NaOH used(L)

=(vol in ml)*(1L/1000ml)

0.00778L 0.00792L 0.00741L

11)mol of KHP titrated

KHP+NaOH-->NaKP+H2O(1:1 molar ratio of KHP and NaOH)

mol of KHP=mass of KHP/molar mass of KHP

=0.527g/204.22g/mol

=0.00258 mol

=0.5168g/204.22g/mol

=0.00253 mol

=0.5143g/204.22g/mol

=0.00252 mol

12) mol of NaOH required for titration(=mol of KHP) =

=0.00258 mol

=0.00253 mol 0.00252 mol 13)molarity of NaOH(mol/L) (mol of NaOH/volume used for titration)

0.00258mol/0.00778L

=0.332 mol/L

==0.00253mol/0.00792L

=0.319mol/L

0.00252 mol/0.00741L

=0.340 mol/L

14) mean molarity of NaOH(mol/L) (0.340+0.319+0.332)/3=0.330 mol/L

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