Question

Your instructor has asked you to prepare 2.00 L of 0.169 M NaOH from a stock solution of 53.4 (+/- 0.4) wt % NaOH with a density of 1.52 (+/- 0.01) g/mL. (a) How many mL of stock solution will you need? (b) If the uncertainty of delivering NaOH is +/- 0.10 mL, calculate the absolute uncertainty in the molarity (0.169 M). Assume negligible uncertainty in the molecular weight of NaOH and in the final volume, 2.00 L.

Answer 1

First we will calculate the molarity of stock solution

We know that molarity = Moles of NaOH / Volume in litres

                                            = Mass of NaOH / Molecular weight of NaOH X Volume of Solution

Mass / Volume = Density = 1.52 (+/- 0.01) g/mL

so molarity =[ 53.4 (+/- 0.4) / 100 X 1000 mL X Density ] / Molecular weight of NaOH

molarity = 20.292 Molar +/- 0.14

New moalrity required = 0.169 M

Volume = 2Litres

M1V1 = M2V2

0.169 X 2 = 20.292 X V2

V2 = 0.0167 L = 167 mL +/- 0. 5

Absolute uncertainity in the molarity = 0.51