Sample Calculations:
Volume of 6 M HCl needed to be diluted to 1 L to make 1 L of 0.1 M HCl
VCon =
Vd Md 1 L×0.10 M
=
=1.67×10-2 L
MCon
6.0 M
0.0167 L×
1000 mL
=16.7 mL
1L
Amount of pure Na2CO3 required to completely react with 25 mL of 0.1 M HCl
0.025 L×
0.10 mol HCl 1 mol Na2 CO3 105.989 g
×
×
Na2 CO3 =0.13 g
1L
2 mol HCl
1 mol
M HCl = mass Na2CO3(g)×
1 mol
2 mol HCl
1
Na2CO3×
×
105.989 g
1 mol Na2CO3 Ve HCl (L)
M HCl=0.1350 g Na2 CO3 ×
1 mol
2 mol HCl
1
Na2 CO3 ×
×
105.989 g
1 mol Na2 CO3 0.02617 L HCl
=0.09734 M HCl [Trial 1]
Amount of impure Na2CO3 (30.00 wt%) required in order to titrate 25.00 mL of 0.1 M HCl
0.025 L×
0.10 mol 1 mol Na2 CO3 105.989 g
100 g Unknown
×
×
Na2 CO3 ×
=0.44 g
1L
2 mol HCl
1 mol
30 g Na2 CO3
mol
1 mol Na2 CO3 105.989 g
Ve of HCl (L) × Conc( L HCl)×
×
Na2CO3
2 mol HCl
1 mol
Mass % Na2 CO3 =
×100%
Mass of Unknown Sample (g)
Mass % Na2 CO3 =
0.02518 L HCl×
0.09748 mol
1 mol Na2 CO3 105.989 g
HCl×
×
Na2 CO3
1L
2 mol HCl
1 mol
×100%
0.4419 g Unknown Sample
=29.43% Na2 CO3 [Trial 1]
0