estion 6 of 6 (1 point) COAST Tutorial Problem A 125.0 mg sample of an unknown,

Question

estion 6 of 6 (1 point) COAST Tutorial Problem A 125.0 mg sample of an unknown, monoprotic acid was dissolved in 100.0 mL of distilled water and titrated with a 0.050 M solution of NaOH. the pH of the solution was monitored throughout the titration, and the data were collected in the table below. Determine the K& of the acid and use Appendix 5 to identify the unknown acid. Volume of OH pH Number Added (mL) 3.09 3.65 4.10 4.50 4.55 Identity of acid: 18 4.94 20 5.11 5.37 5.93 6.24 Check Answer View S Hint

Explanation / Answer

Lets represent the acid by HA and base by BOH. Together they form the salt AB.

At the equivalence point, there is a sudden increase in pH. By looking at the data, we find that at after addition of 22.2 mL base, the pH changes suddenly from 6.24 to 9.91 after further addition. So we take the pH at equivalence point to be the average of these values.

So, pH at equivalence point = (6.24+9.91)/2 = 8.08

Volume of base used at equivalence point = (22.2+22.6)/2 = 22.4 mL

So, at the point of half equivalence, volume of base used = 22.4/2 = 11.2 mL

After addition of this much base, pH recorded is roughly 4.2

Since pH at half equivalence is equal to pKa, so for this acid, pKa = 4.2

So, Ka = 10^-4.2 = 6.3*10^-5

Next, moles of base used at equivalence point = Volume * Molarity = 0.0224 * 0.05 = 0.00112 moles

Since acid is monoprotic, so moles of acid present is also equal to 0.00112

So, MW of acid = Mass/moles = 0.125/0.00112 = 111.6 g

Potassium hydrogen oxalate is the identity of the acid since it has the closest data when looking at appendix.

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