Question

Help! I'm not quite sure how I should set up this experiment (as in how much I should weigh of the dry powder, etc)

The question asks:

" Given the materials list below and Table 2, design a procedure to make the three stock solutions in sufficient amounts to complete the experiment with minimal waste.

Solutions: 0.10 M ascorbic acid, 4.0x10^-4 M methylene blue, 1.20 M HCl

Materials: Ascorbic acid (dry powder); methylene blue (supplied as a hydrated chloride salt—3,7-bis(dimethylamino)phenazathionium chloride —C₁₆H₁₈ClN₃S • 3H₂O); 6 M HCl, analytical balance, three 100 mL volumetric flasks; a 10 mL graduated cylinder, DI water "

Table 2

Set A (4 Solutions)		Set B (4 solutions)		Set C (4 solutions)		Set D (3 solutions)
5.00 mL HCl 5.00 mL Asc		5.00 mL HCl 2.00 mL MB+		10.00 mL Asc 2.00 mL MB+		2.00 mL MB+
H2O (mL)	MB+ (mL)	H2O (mL)	Asc (mL)	H2O (mL)	HCl (mL)	H2O (mL)	Asc (mL)
13	2.00	17	1.00	12	1.00	18	5.00
11	4.00	16	2.00	11	2.00	18	5.00
9	6.00	13	5.00	10	3.00	18	5.00
7	8.00	8	10.00	9	4.00

Answer 1

Total Ascorbic acid needed=5+18+10+15 mL=48 mL

Total MB+needed=20+2+2+2=26 mL

Total HCl=5+5+10=20 mL

Total H2O=40+54+42+54=190 mL

Ascorbic acid =176.12g/mol molecular weight.

50 mL 0.1 M is good enough.

So,50×0.1×176.12/1000=0.88 g needed and 50 mL water can be added into it.

Methylene blue chloride=373.85 g/mol

50 mL 4× 10^-4 M methylene blue=0.007 g needed.50 mL water can be added.

1.2 M HCl needed of 20 mL needed.stock=6 M

6×x=1.2×20

x=4 mL of 6M HCl and 20-4=16 mL water added to nake 1.2 M HCl.