Help! I am completely lost on this chem problem! I still do not know how to work with two different endpoints?

Question

My answer is 36.95% Na2co3 and 47.29%NaHCO.... is this right?

A sample weighing 0.5375 g and containing Na2CO3 (106.0 g/mol), NaHCO3 (84.01 g/mol), and inert material is dissolved and titrated with 0.1014 M HCl. The first end point volume (phenolphthalein end point) is 13.41 mL and the second end point volume (bromocresol green end point) is 69.65 mL. Calculate the weight percentage of Na2CO3.

Answer 1

Actually I am afraid CH4 is not completely right.

During the first titration you will have only the reaction
Na2CO3 + HCl -> NaHCO3 + NaCl

For titration the rule is greq(acid) =greq(base)

and greq = (number of H+ or OH- )* mole

In this reaction you have HCl acid giving 1 H+ and Na2CO3 acting as a monoprotic base since it accepts only 1 H+.

Therefore 1*mole(HCl) =1*mole(Na2CO3)
but mole(HCl)= M(HCl)*V1
and mole Na2CO3 = m(Na2CO3) / MW(Na2CO3)
so

M(HCl)*V1=m(Na2CO3) / MW(Na2CO3) =>
0.1014*0.01341 =m(Na2CO3)/ 106 =>
m(Na2CO3)= 0.1441 g

*** Note: be careful with the units. If you had volume on both sides of the equation then you would not have to do the conversion ml to liters (the factor for conversion would be simplified). Here however we have volume only on one side of the equation and thus we must make sure that everything is expressed appropriately.

Your sample has also inert material so the percentage of Na2CO3 is equal to m(Na2CO3)/m(sample)*100% =
=26.81%

You've already have this answer from funkyourcouch in your other post.

If the sample was only Na2CO3 and NaHCO3 you wouldn't really need a second titration because then NaHCO3 percentage would be 100-Na2CO3 percentage.

In the second titration you have two reactions

Na2CO3 + 2HCl -> H2CO3 + 2NaCl and
NaHCO3 + HCl -> H2CO3 + NaCl

again you have greq(acid)=greq(base)
but now you have 2 bases and this time Na2CO3 acts as a diprotic base.
so 1*mole(HCl)= 2*mole(Na2CO3) + 1*mole(NaHCO3)

we already know the mole(Na2CO3) from the first titration.
is equal to M(HCl)*V1 so

1*M(HCl)*V2 = 2*M(HCl)*V1 + m(NaHCO3) / MW(NaHCO3) =>

m(Na2CO3) = MW(Na2CO3)* M(HCl)* (V2-2V1) =
= 84 * 0.1014*(0.06965-2*0.01341)
= 0.3648 g

so the percentage is 0.3648/0.5375 *100%= 67.87%

and the inert material is 100-(26.81+67.87)= 5.32%

Answer 2

Mw Nahco3

Answer 3

when yo start adding HCl, first all of Na2CO3 will react and form NaHCO3.
then all of the NaHCO3 will react and form H2CO3.
so 13.41ml*2=26.82ml of the total 69.65ml HCl is used for Na2CO3 and the remaining 42.83ml is for NaHCO3. so we can say:
M(Na2CO3)*V(volume of the sample)=0.1014*26.82
M(NaHCO3)*V(volume of the sample)=0.1014*42.83
so we have:
M(Na2CO3)/M(NaHCO3)= 0.6262 ->
W1(weight of Na3CO3)/W2(weight of NaHCO3)=
0.6262*84.01/106.0=0.4963
%Na2CO3=W1/(W1+W2)
=(0.4963W2)/(1.4963W2)
%Na2CO3=(0.4963/1.4963)*100
%Na2CO3=33.17%
%NaHCO3=66.83%

Answer 4

Methane's got it right. Beat me to it.