APPENDIX – List of Figures

Figure 1. Graph of ph against equivalents of HCl titrant

APPENDIX – List of Tables

Table 1. The volume relationships for different compositions of soda ash

APPENDIX – Calculations

Standardization of Hydrochloric Acid Solutiona. Trial 1Net volume of HCl

(37.7±0.10 )− (0±0.10 )=37.7mL±0.14mLR=37.7mL

r=√ (0.10 )2+ (0.10 )2=0.141421356

M HCl

0.1037 g±0.0002g Na2C O3×1mol Na2CO3

106 g Na2CO3

×2mol HCl

1mol Na2C O3

×1

37.7mL±0.14mL×

1000mL1 L

=0.0519M ±0.0002

R=0.1037 gNa2CO3×1mol Na2CO3

106 g Na2CO 3

×2mol HCl

1mol Na2CO3

×1

37.7mL×

1000mL1 L

=0.0519042010M

r=R √( 0.00020.1037 )

2

+( 0.1437.7 )

2

=0.000218931015

b. Trial 2Net volume of HCl

(38.8±0.10 )−( 0±0.10 )=38.8mL±0.14mLR=38.8mL

r=√ (0.10 )2+ (0.10 )2=0.141421356

M HCl

0.1052 g±0.0002 g Na2CO3×1mol Na2CO3

106g Na2CO3

×2mol HCl

1mol Na2CO3

×1

38.8mL±0.14mL×

1000mL1L

=0.0512M±0.0002

R=0.1052g Na2C O3×1mol Na2CO3

106 g Na2CO3

×2mol HCl

1mol Na2C O3

×1

37.7mL×

1000mL1 L

=0.0511573624M

r=R √( 0.00020.1052 )

2

+( 0.1438.8 )

2

=0.000210302736

c. Trial 3Net volume of HCl

(36.5±0.10 )−(0±0.10 )=36.5mL±0.14mLR=36.5mL

r=√ (0.10 )2+ (0.10 )2=0.141421356

M HCl

0.0989 g±0.0002g Na2CO3×1mol Na2C O3

106 gNa2CO3

×2mol HCl

1mol Na2CO3

×1

36.5mL±0.14mL×

1000mL1 L

=0.0511M ±0.0002

R=0.0989g Na2C O3×1mol Na2CO3

106 g Na2CO3

×2mol HCl

1mol Na2C O3

×1

36.5mL×

1000mL1 L

=0.0511243215M

r=R √( 0.00020.0989 )

2

+( 0.1436.5 )

2

=0.000223441191

Average M HCl(0.0519±0.0002 )+ (0.0512±0.0002 )+(0.0511±0.0002)

3=0.0514M ±0.0003

r=√0.00022+0.00022+0.00022=0.000346410162

Percentage composition of the samplea. Trial 1

%Na2CO3=(M HCl (ave ) ) (V ph )(FW Na2CO3

)

g of sample×100 %

%Na2CO3=(0.0514M ±0.0003 ) (13.9mL±0.14mL)( 1L

1000mL )( 106g Na2CO3

1mol Na2CO3

)

0.5079g±0.0002 g×100%

Rnumerator=0.0757327600

rnumerator=R√( 0.00030.0514 )

2

+( 0.1413.9 )

2

=0.000881594525

0.0757327600±0.00090.5079 g±0.0002g

×100 %

R=0.149109589

r=R √( 0.00090.0757 )

2

+( 0.00020.5079 )

2

=0.00173436853

%Na2CO3=14.91 %±0.0017

%NaHCO3=(MHCl (ave ) ) (V mo−V ph )(FW NaHCO 3

)

gof sample×100 %

%NaHCO3=(0.0514M ±0.0003 ) ( (19.2mL±0.14 )−(13.9mL±0.14))( 1 L

1000mL)(84.01 g NaHCO3

1mol NaHCO3

)

0.5079 g±0.0002 g×100%

RV mo−V ph=5.3mL

rV mo−V ph=√0.142+0.142=0.197989899Rnumerator=0.0228860042

rnumerator=R√( 0.00030.0514 )

2

+( 0.1985.3 )

2

=0.000865314958

0.0228860042±0.0008653149580.5079 g±0.0002 g

×100 %

R=0.0450600595

r=R √( 0.00090.0229 )

2

+( 0.00020.5079 )

2

=0.00170380367

%NaHCO3=4.51% ±0.0017

b. Trial 2

%Na2CO3=(M HCl (ave ) ) (V ph )(FW Na2CO3

)

g of sample×100%

%Na2CO3=(0.0514M ±0.0003 ) (13.1mL±0.14mL)( 1 L

1000mL )(106 g Na2CO3

1mol Na2CO3

)

0.5079g±0.0002g×100 %

%Na2CO3=14.05 % ±0.0017

%NaHCO3=(MHCl (ave ) ) (V mo−V ph )(FWNaHCO 3

)

gof sample×100%

%NaHCO3=(0.0514M ±0.0003 ) ( (18.3mL±0.14 )−(13.1mL±0.14))( 1 L

1000mL )(84.01 g NaHCO3

1mol NaHCO3

)

0.5079 g±0.0002 g×100 %

%NaHCO3=4.42% ±0.0017

c. Trial 3

%Na2CO3=(M HCl (ave ) ) (V ph )(FW Na2CO3

)

g of sample×100%

%Na2CO3=(0.0514M ±0.0003 ) (14.3mL±0.14mL)( 1L

1000mL )( 106g Na2C O3

1mol Na2CO3

)

0.5079g±0.0002 g×100 %

%Na2CO3=15.34 %±0.0017

%NaHCO3=(MHCl (ave ) ) (V mo−V ph )(FWNaHCO 3

)

gof sample×100%

%NaHCO3=(0.0514M ±0.0003 ) ( (18.9mL±0.14 )−(14.3mL±0.14))( 1 L

1000mL )( 84.01g NaHCO3

1mol NaHCO3

)

0.5079 g±0.0002 g×100 %

%NaHCO3=3.91 %±0.0017

Determination of Vph and Vmo

Vph=∑ net volume of HCl at phenolphthalein endpoint

3=

13.9+13.1+14.33

= 13.77

Vmo= ∑ net volume of HCl at methyl orange endpoint

3 =

19.2+18.3+18.93

= 18.80