[13%] An unknown monoprotic acid contains only the elements C, H, and O. — Titration and Analytical Chemistry Chemistry Question
Problem Context
[13%] An unknown monoprotic acid contains only the elements C, H, and O.
Combustion analysis indicates that the unknown acid contains 40.0% C and 6.7% H by mass. What is its empirical formula?
Model Answer
1.000 g of the acid would contain 0.400 g C, 0.067 g H, and (by difference) 0.533 g O.
0.400 g C/(12.01 g mol-1) = 0.0333 mol C
0.067 g H/(1.008 g mol-1) = 0.0665 mol H
0.533 g O/(16.00 g mol-1) = 0.0333 mol O
So the empirical formula is CH2O.
A 1.000 g sample of the compound is dissolved in 50 mL water and titrated with 0.3000 M NaOH to a phenolphthalein endpoint, which is observed after the addition of 37.00 mL of base. What is the molar mass of the acid and what is its molecular formula?
Model Answer
0.03700 L titrant 0.3000 mol L-1 = 0.01110 mol NaOH = 0.01110 mol unknown
1.000 g unknown/0.01110 mol unknown = 90.09 g mol-1
Since the formula mass of CH2O is 30.03 g mol-1, there must be three formula units per molecule; the molecular formula is thus C3H6O3.
The pH of the titration mixture is measured to be 3.43 after the addition of 10.00 mL of the NaOH solution. What is the pKa of the unknown acid?
Model Answer
After 10.00 mL base have been added 10.0/37.0 of the acid has been deprotonated, while 27.0/37.0 remains protonated. Using the Henderson-Hasselbalch equation:
pH = pKa + log10([X –]/[HX])
3.43 = pKa + log10(10/27)
pKa = 3.86
A different monoprotic acid with the same molar mass is titrated under the same conditions and the pH of the solution is measured after the addition of 40.00 mL NaOH solution. Explain why this measurement will not be informative about the pKa of this acid, and estimate the pH at this point in the titration.
Model Answer
Since this acid is also monoprotic and has the same molar mass, the endpoint will also be at 37.00 mL added base. Thus, at 40.00 mL added NaOH, essentially all the original acid will have been deprotonated, and the pH will be determined by the amount of excess NaOH added.
Specifically, 3.00 mL excess base will have been added
0.00300 L 0.3000 mol L-1 = 9.00 10-4 mol excess OH–
Total volume = 50 mL original + 40 mL added = 90 mL
[OH–] = 9.00 10-4 mol/0.090 L = 1.0 10-2 M
pH = 14.0 + log10[OH–] = 12.0
An aqueous solution of the unknown acid was found to rotate the plane of polarization of plane-polarized light (at the sodium D line). Suggest a reasonable structure for the unknown acid.
Model Answer
Since a solution of the unknown acid rotates plane-polarized light, the acid must be chiral. An organic acid with a pKa = 3.86 and a formula of C3H6O3 must contain a carboxylic acid, –COOH. The only chiral carboxylic acid with this formula is lactic acid, CH3CH(OH)COOH, though of course one cannot determine which enantiomer is present from the given data.