Question

DATE SECTION - B. Analysis of Vinegar average molarity of NaOH (from part A), M HC,H,O, (aq) + NaOH(aq) → H,O (1) + NaC,H,O, (aq) Trial 1 Trial 2 Trial 3 volume of vinegar, ml 10mL 10mL 10mL volume of vinegar, L initial buret reading, mL final buret reading, ml 0.25mL 0.11mL 0.32mL 30.01ML 31.5mL 28.50mL volume of NaOH used, mL volume of NaOH used, L moles of NaOH, mol moles of HC,H,O, (aq), mol__ [HC,H,O, (aq)], M average (HC,H,O, (aq)], M Show calculations and calculate the mass percent concentration of HC,H,O, on the back of this page. REPORT SHEET

the volume, moles and average need to be calculated from the calculated numbers i got above

the molarity of NaOH is 0.2998M

Answer 1

1) Volume of vineger (in L)

For all trails : volume = 10/1000 L = 0.01 L

2) Volume of NaOH (ml) = final buret reading - initial reading

For trail 1 : volume of NaOH = 30.01-0.25 = 29.76 ml

For trail 2 : volume of NaOH = 31.5-0.11 = 31.39 ml

For trail 3 : volume of NaOH = 28.50-0.32 = 28.18 ml

3) Volume of NaOH in litres :

For trail 1 : volume in (L) = 29.76/1000 = 0.02976 L

For trail 2 : volume in (L) = 31.39/1000 = 0.03139 L

For trail 3 : volume in (L) = 28.18/1000 = 0.02818 L

4) Moles of NaOH = molarity×volume(in L)

For trail 1 : Moles of NaOH = 0.2998×0.02976 = 0.008922

For trail 2 : moles of NaOH = 0.2998×0.03139 = 0.009411

For trail 3 : moles of NaOH = 0.2998×0.02818 = 0.008448

5) Moles of HC2H3O2 = moles of NaOH

For trail 1 : moles of HC2H3O2 = 0.008922

For trail 2 : moles of HC2H3O2 = 0.009411

For trail 3 : Moles of HC2H3O2 = 0.008448

6) [HC2H3O2] = moles×1000/volume(ml)

For trail 1 : [HC2H3O2] = 0.008922×1000/10 = 0.8922 M

For trail 2 : [HC2H3O2] = 0.009411×1000/10 = 0.9411 M

For trail 3 : [HC2H3O2] = 0.008448×1000/10 = 0.8448 M

7) Average Molarity = (0.8922+0.9411+0.8448)/3

= 0.8927 M