Question

From one of the documents, Determination of Ka of a weak acid Version Version 42-0151-00-02, they got the answers Data5 Table 3. Determination of Unknown pKa of Unknown Weak Acid: 4.75 Ka of Unknown Weak Acid: 1.76 x 10-5 % error pKa: 0% % Error Ka: 1.67% Can someone show work on how they got this?

Drops NaOH Added	pH Value Trial 1	pH Value Trial 2	pH Value (Average)
Half-Equivalent Point = (53 Drops)	4	5	4.5
Equivalent Point = (106 Drops)	8	8.5	8.5
0	2	2	2
10	2	3	2.5
20	3	3	3
30	4	4	4
40	4	4	4
50	4	5	4.5
60	6	6	6
70	6	7	6.5
80	7	7	7
90	7	7	7
100	8	7	7.5
110	8	8.5	8.5
120	13	13	13

Answer 1

This is all about making a graph, here´s a graph for your ph value at half equivalence point = 5

you know where is your equivalence point, now you need to determine where is your half eequivalence point, it is at the half of it

I´m using your value of 53 drops (approximately because it is a graph method)

you have to read the ph from the half equivalence point which is 5

at half equivalence point PH = PKa, you can repeat this same procedure for the other graph , you already have that info the ph at 50 drops is 4 and 5, that´s why pka are 4 and 5 for this trials

Remember that PKa = - log (ka)

so the exact value is 4.75 for pka

ka = 10^-pka = 10^-4.75 = 1.778 x10^-5

Ka for experiment (from average ph = 4.5)

ka = 10^-pka = 10^-4.5 = 3.162 x10^-5

% error is ( theoretical value - actual value) / theoretical value * 100

pka experimntal average = 4.5

Pka theoretical = 4.75

% error = (4.75 - 4.5) / 4.75 = 0.052 * 100 = 5.2% error for pka

similarly for ka you will get a value of 77.8% error

this constrast your information but this is the formula, it depends on the values you are trying to compare.

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