When a 3.529 g sample of Liquid-Plumr was titrated with 0.1025 M HCl two end points were obtained. The first end point occurred at 16.33 mL the second at 41.58 mL. Calculate the mass percent of both N?

Answer 1

Liquid-Plumr is a chemical drain opener made of 0.5-2% sodium hydroxide (NaOH, M=40,00 g/mol) and 5-10% sodium hypochlorite (NaClO, M=74.44 g/mol), and a surfactant (detergent, unknown), according to the producer Clorox (to calculate mass% these underscored date are essential)

• The first part of this titration is for NaOH, being the strongest alkali:

0.1025(mmol/mL)*16.33(mL)*40.00(mg/mmol)*0.001(g/mg) / 3.529(g) * 100% = 1.897% NaOH

• The second part is for NaClO, a weak, though still titrable base when the right indicator is used.

At first it is assumed that the total volume was 41.58 mL, so the second part of this titration only took (41.58-16.33) = 25.25 mL (net value), then

0.1025(mmol/mL)*(25.25) (mL)*74.44(mg/mmol)*0.001(g/mg) / 3.529(g) * 100% = 5.458% NaClO

(this looks this most reasonable value to me, because the tatration can be done with one buret filling)

However, if 41.58 mL is used to titrate the second part (thus total volume would have been 59.91) then:

0.1025(mmol/mL)*(41.58)(mL)*74.44(mg/mmol)*0.001(g/mg) / 3.529(g) * 100% = 8.988%

(I would advise to use a smaller weight of sample next tim, e.g. about 2 gram LP)

Note: all value were calculated in 4 signifcant digits according to the given figures.

However this does not suggest that the outcomes are as accurate; at least triplicate titrations on 3 different weights (amounts) of the same sample are needed for an accuracy of <0.1% of outcome values.