Chem 26.1 Calculations ATQ_6

D. Gevaña/ Chemistry 26.1 (2017)

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CALCULATIONS 1. molarity of the standard HCl solution and report it as M

±

∆ M 

 Trial  Trial 1 Vinitial Cl ! (0 " 0.0#) m$

0.05 + 0.05 = 0.0707106 0.0707106 7812 ≈ 0.07 mL √ 0.05 2

%V !

2

V&nal Cl ! ('.6 " 0.0#) m$ Vnet Cl ! '.6 m$

Cl !

mole mole HCl HCl  x g Na2 CO 3 x purity mole Na2 CO 3  ! mL HCl HCl x MW of of Na2 CO3

2 mol mol HCl HCl

mol Na2 CO 3 1 mol

 x 0.1012 g Na2 CO3 x 0.999

 !

g Na2 CO 3 0.0386  L x 105.988 mole

0.0*+*2''2++ ! 0.0*+*2  ∆ M =¿ 0.0*+*2''2++ 0.0*+*2''2++  Cl ,

0.0001 

√(

0.0002 0.1012

)( ) 2

+

0.07

2

38.6

 ! 1.'2#6*+'7' , 10-*

≈

 M  HCl ± ∆ M  HCl =( 0.04942 ± 0. 0001 0001 )  M HCl

 Trial  Trial 2 Vinitial Cl ! (0 " 0.0#) m$

0.05 + 0.05 = 0.0707106 0.0707106 7812 ≈ 0.07 mL √ 0.05 2

%V !

2

V&nal Cl ! ('.' " 0.0#) m$ Vnet Cl ! '.' m$

Cl !

mole mole HCl  x g Na2 CO 3 x purity mole mole Na2 CO 3  ! mL HCl HCl x MW of of Na2 CO3

HCl 2 mol HCl mol Na2 CO 3 1 mol

 x 0.1010 g Na2 CO3 x 0.999

 !

g  Na 2 CO 3 0.0386 L x 105.988 mole

0.0*+'2#6#**2 ! 0.0*+''  0.0*+'2#6#**2  Cl , ∆ M =¿ 0.0*+'2#6#**2 0.0001 

√(

0.0493 3 ± 0. 0001 0001 ) M HCl  M  HCl ± ∆ M  HCl =( 0.0493

0.0002 0.1010

)( ) 2

+

0.07 38.3

2

 ! 1.'2+1+*#6 , 10-*

≈

D. Gevaña/ Chemistry 26.1 (2017)

!

Ave. MHCl

∆ M   ! 0.0002 

0.049 42332899

+ 0.049 32565442 2

=¿  0.0*+'7**+171 

√ ( 0.0001 325649373) +( 0.0001329194568 ) 2

2

Ave.  M  HCl ± ∆ M  HCl ! .!"#$

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 ! 1.772#+27 , 10-*

≈

± 0.0002  M HCl

%. per&enta'e &omposition of the sample and report it as (A

±

∆ A

Trial 1 a. Na%CO#

Vinitial Cl ! (0 " 0.0#) m$ V&nal Cl ! (16.1 " 0.0#) m$ Vnet Cl (henlhthalein enint) ! 16.1 m$  %V !

√ 0.05 + 0.05 =0.07071067812 ≈ 0.07 mL 2

2

Vh ! (16.1 " 0.07) m$ ! (0.0161 " 0.00007)$

 !

 Ave M  HCl x V  pH  x MW of Na2 CO3 gof sodaash

0.04937449171 M x 0.0161 L x 1 05.988 0.5003

∆ A =0.0842529684  g x

g Na2 CO3

 !

gof sodaash

g  Na 2 CO3 mole

g of soda ash

√(

  0.0002 0.04937

) +( 2

0.00007 0.0161

)= 2

0.0005006820266 ≈ 0.00050

g

 A ± ∆ A =( 0. 08425 ± 0.0005 ) g (A )

0.04937449171 M x 0.0161 L x 1 05.988 0.5003 g of

(A *

∆ A  ) 1+.,! (

-. NaHCO#

Vinitial Cl ! (16.1 " 0.0#) m$

soda ash

± 0.0005

g Na2 CO 3 ! 0.16*0*+'+ ! 16.*  mole

D. Gevaña/ Chemistry 26.1 (2017)

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V&nal Cl ! (*2.' " 0.0#) m$ Vnet Cl (methyl range enint) ! 26.2 m$  %V !

√ 0.05 + 0.05 =0.07071067812 ≈ 0.07 mL 2

2

Vm ! (26.2 " 0.07) m$ ! (0.0262 " 0.00007) $ VCl ! Vm - V ! 26.2 m$ 3 16.1 m$ ! 10.1 m$ ! 0.0101 $  %V !

√ 0.07 +0.07 =0.0 9899494937 ≈ 0.10 mL 2

2

VCl ! (10.1 " 0.10) m$ ! (0.0101 " 0.0001) $

 !

 Ave M  HCl x V  pH  x MW of NaH CO3 gof sodaash

0.04937449171 M x 0.01 01 L x 84.008 0.5003 g of

∆ A =0.0 4189330823 g x

√(

g NaH CO3

 !

gof sodaash

g  NaH CO3 mole

soda ash

) +( 2

  0.0002 0.04937

0.0001 0.0101

)= 2

0.000 4481615934 ≈ 0.0004

g

 A ± ∆ A =( 0.0 4189 ± 0.0004 ) g (A )

0.04937449171 M x 0.0161 L x 1 05.988 0.5003 g of

soda ash

 (A *

∆ A  ) ,.#$ (

± 0.000 4

Trial % a. Na%CO#

Vinitial Cl ! (0 " 0.0#) m$ V&nal Cl ! (16.1 " 0.0#) m$ Vnet Cl (henlhthalein enint) ! 16.1 m$  %V !

√ 0.05 + 0.05 =0.07071067812 ≈ 0.07 mL 2

2

Vh ! (16.1 " 0.07) m$ ! (0.0161 " 0.00007)$

g Na2 CO 3 ! 0. 0'7'6'7*6' ! .'7 mole

D. Gevaña/ Chemistry 26.1 (2017)

 !

 Ave M  HCl x V  pH  x MW of Na2 CO3 gof sodaash

0.04937449171 M x 0.0161 L x 1 05.988 0.5003

∆ A =0.0842529684  g x

Page | *

g Na2 CO3

 !

gof sodaash

g  Na 2 CO3 mole

g of soda ash

√(

  0.0002 0.04937

) +( 2

0.00007 0.0161

)= 2

0.0005006820266 ≈ 0.00050

g

 A ± ∆ A =( 0. 08425 ± 0.0005 ) g (A )

0.04937449171 M x 0.0161 L x 1 05.988 0.5003 g of

(A *

∆ A  ) 1+.,! (

g Na2 CO 3 ! 0.16*0*+'+ ! 16.*  mole

soda ash

± 0.0005

-. NaHCO#

Vinitial Cl ! (16.1 " 0.0#) m$ V&nal Cl ! (*1.7 " 0.0#) m$ Vnet Cl (methyl range enint) ! 2#.6 m$  %V !

√ 0.05 + 0.05 =0.07071067812 ≈ 0.07 mL 2

2

Vm ! (2#.6 " 0.07) m$ ! (0.02#6 " 0.00007) $ VCl ! Vm - V ! 2#.6 m$ 3 16.1 m$ ! +.# m$ ! 0.00+# $  %V !

√ 0.07 +0.07 =0.0 9899494937 ≈ 0.10 mL 2

2

VCl ! (+.# " 0.10) m$ ! (0.00+# " 0.0001) $

 !

 Ave M  HCl x V  pH  x MW of NaH CO3 gof sodaash

0.04937449171 M x 0. 0095 L x 84.008 0.5003 gof

∆ A =0.0 3940459685 g x

√(

g NaH CO3

 !

gof sodaash

g  NaH CO3 mole

sodaash

) +( 2

  0.0002 0.04937

0.0001 0.0095

)= 2

0.000 44 44417117 ≈ 0.0004

g

 A ± ∆ A =( 0.0 3094 ± 0. 0004 ) g (A )

0.04937449171 M x 0.0161 L x 1 05.988 0.5003 g of



soda ash

g Na2 CO 3 ! 0. 07761+'6#' ! 7. mole

D. Gevaña/ Chemistry 26.1 (2017)

∆ A  ) $.,, (

(A *

± 0.000 4

16.84 + 16.84

Aver ageof%4a2C5' =

2

Aver ageof% 4aC5' =

Page | #

8.37

+ 7.88 2

=16.84

=8.125

3. elative

standard deviation /in ppt0 and &onden&e limits /"2( &onden&e level0 2

−16.84  ¿ ¿ ¿ 2− 1 16.84 −16.84  ¿ +¿ ¿ ¿ √ ¿ 16.84

D 4a2C5' !

2

8D! 0  t

C$!

 x´ ±

ts  ! 16.*  √ n

±

( 12.7)( 0 ) √ 2

 ! 16.* 

9nterval:   4.36790214 − 4.241289226 2

−8.125 ¿ ¿ ¿ 2−1 8.37 −8.125  ¿ +¿ ¿ ¿ √ ¿ 7.88

D 4aC5' !

8D!

C$!

  0.3464823228 8.125

 x´ ±

2

.'*6*2'22

 1000 ppt   ! *2.6*'+71+ t

ts  ! .12#  √ n

9nterval:   5.0135 −11.2365

±

( 12.7)( 0.3464823228 ) √ 2

 ! .12# 

± '.111#

D. Gevaña/ Chemistry 26.1 (2017)

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