Question

I found the PKa1 values through experiment

could you show me how to get the percent errors and the percent for organic acid contents

pretty much all the empty spaces. i have no idea where to start

General Chemistry Il Laboratory / 10 points Experimental Data and Results Data Table 2: Titration of Acids in Juice MeasureNet Station No. N/A Juice Sample Apple Cranberry (Monoprotic) 24.9275 Mass of Sample (9) Diprotic) 24. 6864 Initial Buret Reading (mL) 0.00 0.00 Final Buret Reading (ml) 28.87 49.80 49.80 Volume of NaOH to Eq. Pt. (mL) 28.87 Molarity of NaOH (mol/L) 0.1000M Moles of NaOH to Eq. Pt. (mol) .04980 M .02887M Experimental pal (from fitration curve) 3.244 3.95 Literature Value of pkai (from Data Table 1) 4.33 3.40 Experimental pk. (from Hitration curve) 4.94 Literature Value of pk.2 (from Data Table 1) 5.20 Percent Error for pk 1 (%) Percent Error for pk. (%) Organic Acid Content (g/100 g) Percent Error for Organic Acid Content (%) Experiment 7. Determination of the Acid Concentration in Julco 177

Answer 1

ANSWER:

The percent error is calculated with the following formula:

%error experimental theoretically 1000 Xtheoretical

Then to calculate the percent error of pKa values, we use the experimental and theoretical (literature values) values of pKa for each juice:

Juice	pKa	Exp pKa	Theor pKa	% error
Cranberry	pKa1	3.244	4.33	13.244-4.33、100% 4.33 25.08%
Apple	pKa1	3.95	3.40	3.95 – 3.40 " x 100% = 16.18% 3.40
pKa2	4.94	5.20	4.94 – 5.20 " x 100% = 5.00% 5,20

For the organic acid content, we need to find the moles of organic acid in the juice (using the titration data) and then, we transform the moles into g/100 g.

• Cranberry juice:

In the titration, you use 49.80 mL (49.80x10-3 L) of NaOH 0.1000 M, then the moles of NaOH used are:

molesNaoh = [NaOH] XVNaoh = 0.1000", -X49.80.c10-3L = 4.98x10-3 mol

The titration of a monoprotic acid has the following reaction

HA + NaOH + A + H2O + Na+

Then, at equivalence poitn, the moles of NaOH used are equal to the moles of organic acid in the juice,

molesNgoh = molesha = 4.98x10-3 mol

In cranberry juice, the principal acid is quinic acid (MW = 192.17 g/mol), then, the amount of quinic acid in the sample (24.9275 g) is:

massha= molesha X MWHA = 4.98x10-3 mol x 192.179,=0.957 g

And the conceten of organic acid (g/100 g juice) is:

0.957 g HA - = 0.03839g/9 juice 24.9275 g juice

in 100 g of juice

3.839 g 100 g juice

• Apple juice:

In the titration, you use 28.87 mL (28.87x10-3 L) of NaOH 0.1000 M, then the moles of NaOH used are:

molesNaoh = [NaOH] XVNaoh = 0.1000", - x 28.87.610-3L = 2.887x10-3 mol L

The titration of a diprotic acid has the following reaction

HA+2 NaOH → A-2 + 2 H20 +2 Nat

Then, at equivalence poitn, the moles of NaOH used are two times the moles of organic acid in the juice,

molesh, A = 2.887.10-3 mol NaOH x; 1 mol H2A = 1.444x10-3 mol *2mol NaOH

In cranberry juice, the principal acid is malic acid (MW = 134.09 g/mol), then, the amount of malic acid in the sample (24.6864 g) is:

massH2A = molesh,AXMW H2A = 1.444x10-3 molx 134.09 9,= 0.194 g

And the conceten of organic acid (g/100 g juice) is:

0.194 g HA 24.6864 g juice = 0.007841g/9 juice

in 100 g of juice

0.7841 g 100 g juice

The percent error in organic acid content is calculated with the same formula than for percent error of pKa. In this case, we use the acid content calculated (experimental value) and the reported by the juice (theorethical value). For example si the cranberry juice reports an organic acid content of 2.50 g/ 100 g and the apple juice reports 0.45 g/ 100 g, then, the percent errors are

Juice	Exp conten	Reported content	% error
Cranberry	3.839 g/100 g	2.50 g/100 g	3.839 – 2.50 I x 100% = 53.56% 2.50
Apple	0.7841 g/100 g	0.45 g/100 g	0.7841 -0.45 " x 100% = 74.24% 0.45

• NOTE: verifiy what are the reported values of organic acid for your samples