Unknown acid .30 o5 (etter or number on unknown bottle) Mass of Unknown acid: Co

Question

Unknown acid .30 o5 (etter or number on unknown bottle) Mass of Unknown acid: Concentration of NaOH (aq) Q, 002 M 1. Plot the titration curve as pH vs. mL of NaoH added using a computer graphing program. Make sure you have chosen the correct x and Y axes. Please include gridlines and print out a 8 's x 11 plot. You must attach the graph to the Lab Report 2. Locate the equivalence point(s) on the titration curve. It is very unlikely that the equivalence point( will be actual data points. You will need to either manually draw a smooth curve connecting y data points, or select the option in your graphing program that will draw the best-fit curve for you. Then find the steepest point(s) on the curve. This is these are the equivalence point(s). Label the equivalence point(s) on the curve. (For polyprotic acids label all the equivaence points.) 3. How many equivalence points does your graph show? 4. Recall that at the first equivalence point, moles of acid moles of base added Volume of NaOH added at the first equivalence point 2 M Moles of NaOH added at the first equivalence point: Volume of NaoH added at the Second equivalence point (if your unknown is polyprotic) 5. Use the mass of the acid, the concentration of NaoH and the volume of NaOH added at the first proper equivalence point to calculate the molar mass of the acid: (Show your work clearly and include units.) 0, 1002 mot NaOH mo H/F) 0.00 Calculated Molar mass of the unknown acide 6. Label the areas of the curve where a buffer is present. Locate the point where pH pKa (halfway to the equivalence point.) Label pKa on the graph. If the acid is bel the pKa for each equivalence point on the graph. (from the graph) 7. Calculate Ka for your unknown acid. f your unknown is polyprotic, calculate Ka at both the equivalence points. (Show work)

Explanation / Answer

4. Volume of NaOH used = 8.5 ml

moles of NaOH = concentration*volume (L) = 0.1002*0.0085 = 0.00085 mol

5. Mass of acid = 0.2065 g

Moles of NaOH used = 0.00085 mol

Moles of acid used = moles of NaOH used = 0.00085 mol

molar mass of acid = mass/moles = 0.2065/0.00085 = 242 g/mol

7. pKa = -log Ka

5.15 = -log Ka

Ka = 7.08*10-8

8. To identify the compound, we know molar mass, which is 242 g/mol and Ka, which is 7.08*10-8. And the acid is monoprotic, as there is only one equivalence point in the graph.

9. Concentration of acid = moles of acid/volume = 0.2065/(242*0.100) = 0.0085 M

10. To find pH of the acid:

HA ----> H+ + A-

Ka = [H+][A-]/[HA]

7.08*10-8 = x2/(0.0085-x)

7.08*10-8 = x2/(0.0085)

x = 2.45*10-4

[H+] = x = 2.45*10-4

pH = -log[H+] = -log(2.45*10-4) = 3.61

pH = 3.61

Experimental pH = 3.94

measured pH was more than the calculated pH.

% error = [(3.94-3.61)/3.94] *100 = 8.38 %

[HA] [H+] [A-] Initial 0.0085 M 0 0 Change -x +x +x Equilibirum 0.0085 -x x x

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