Question

A 100 mg vitamin C tablet weighed 1.627 g. A student crushed the tablet and transferred 0.159 g of the crushed powder into a 100 mL volumetric flask, dissolved the powder in ~30 mL DI water, then filled the flask to the calibration line with DI water and mixed well to create Solution 1. The student then transferred 23.2 mL of Solution 1 into a 100 mL volumetric flask and diluted to the calibration line with DI water to create Solution 2. Calculate the expected concentration (in mM) of ascorbic acid in Solution 2.

Answer 1

1.627 g of the tablet contains 100 mg vitamin C.

So, 0.159 g contains (100 mg / 1.627 g) x 0.159 g = 9.7726 mg.

Molar mass of ascorbic acid is 176 g/mol (C₆H₈O₆ , 6x12 + 8x1 + 6x16 = 176)

So, 9.7726 mg of ascorbic acid = 9.7726 mg / (176 g/mol ) = 0.0555 (mg/g) x mol = 0.0555 mmol.

This much amount was dissolved in 100 mL DI water. So, in 1000 mL mmol of ascorbic acid = (0.0555 mmol/ 100 mL) x 1000 mL = 0.555 mmol/L = 0.555 mM. So, concentration of ascorbic acid in flask 1 was 0.555 mM.

Now, again,

23.2 mL of 0.555 mM solution was diluted into 100 mL in flask 2. So, initial volume = V_initial = 23.2 mL, initial concentration = C_initial = 0.555 mM and final volume = V_final = 100 mL. We know from condition of equivalency,

So, final concentration of ascorbic acid in flask 2 is 0.129 mM.