[bornn]

Hello I need help in calculating % Na2CO3 in the unknown for each of my trials. I really have no idea where to start I did however figure out my first part of my experiment. Well here was the experiment.

The first part of the experiment was to standardization of HCl with Na2CO3.

the average morality of three trials of HCl came to .093298 M HCl

For the second part of the experiment I obtained an vial of unknown soda ash and I used the same procedure for the standardization of HCl except the mass of the unknown was used in place of the primary standard.

For my trials I got the following:

Trial 1
.7893g of soda ash
took 41.11 ml of HCl to neutralize the solution

Trial 2
.6734g of soda ash
took 34.92 ml of HCl to neutralize the solution

Trial 3
.7413g of soda ash
took 38.51 ml of HCl to neutralize the solution

So here comes my question how do I calculate the % Na2CO3 in the unknown for each trial

Any help would be much appreciated, thank you for anybody who takes the time to help.

[fledarmus]

a) How many moles of HCl did you use in each titration?

b) What does that tell you about the amount of sodium carbonate in your sample?

[bornn]

a) How many moles of HCl did you use in each titration?

b) What does that tell you about the amount of sodium carbonate in your sample?

a)Trial 1
   .003174 moles HCl

   Trial 2
   .003066 moles HCl

   Trial 3
   .003174 moles HCl

b)Hmm I think it means it takes 2 HCl to neutralize 1 Na2CO3
since it is a 2:1 ratio according to Na2Co3+2HCl > H2O+CO2+2NaCl

[Borek]

You are almost there. Now, calculate masses of carbonate in each sample, then use percentage definition.

[bornn]

Ok lets see if I am on the right track here

For my trial 1
.7893g and it took 41.12 ML of HCl to neutralize the solution

Mol Na2CO3 =.7893g x 1 mol/106g= .007446 moles

So mass would be. 007446 moles x 106 g
mass equals = .789276

But now do use the ratio to find the percentage sorry hitting a brick wall here

[sjb]

Ok lets see if I am on the right track here

For my trial 1
.7893g and it took 41.12 ML of HCl to neutralize the solution

Mol Na2CO3 =.7893g x 1 mol/106g= .007446 moles

So mass would be. 007446 moles x 106 g
mass equals = .789276

But now do use the ratio to find the percentage sorry hitting a brick wall here

Not quite. If you used 41.13 ML (are you sure about this, check your notation) of HCl (what concentration?), how many moles of acid did you use up? From a balanced equation, how many moles of sodium carbonate does this react with? What is the mass of this amount of sodium carbonate? Hence what percentage of your original sample is sodium bicarbonate.

What you have done is to calculate the mass of sodium carbonate using the mass of sodium carbonate as a starting point, any slight differences are due to rounding off answers and significant figures.

(edit out misleading typo, sjb)

[fledarmus]

Ok lets see if I am on the right track here

For my trial 1
.7893g and it took 41.12 ML of HCl to neutralize the solution

Mol Na2CO3 =.7893g

This isn't the mass of Na2CO3, it is the total amount of ash which you are trying to analyze to see what percentage is Na2CO3. You will need to go the other direction, from the .003174 moles of HCl that you calculated previously. As you also mentioned, it takes 2 moles of HCl to neutralize each mole of Na2CO3. So how many moles of Na2CO3 were in the sample? How much would that much Na2CO3 weigh? and what percentage is that of your total weight of ash?

[bornn]

Ok lets see if I am on the right track here

For my trial 1
.7893g and it took 41.12 ML of HCl to neutralize the solution

Mol Na2CO3 =.7893g

This isn't the mass of Na2CO3, it is the total amount of ash which you are trying to analyze to see what percentage is Na2CO3. You will need to go the other direction, from the .003174 moles of HCl that you calculated previously. As you also mentioned, it takes 2 moles of HCl to neutralize each mole of Na2CO3. So how many moles of Na2CO3 were in the sample? How much would that much Na2CO3 weigh? and what percentage is that of your total weight of ash?

You are correct my mistake that is the mass of soda ash. That .003174 moles of HCl came from one of my trials in the first part of the experiment (the standardization of HCl.

So far this is what i have calculated for the second part of the experiement which is to find the % Na2CO3 in the unknown.

.7893 grams of soda ash
41.12mL of HCl

mol=.7893g X 1mol/106=.007446

l soln=41.12mLX 1L/1000mL=.04112

M=.007446/.04112=.181080 M

.007446mol X 2 molHCl/1mol =.014892 moles HCl

So how do I calculate the percent of Na2CO3 in the unknown?
I feel like this is really trivial but I am just not getting it sorry; for my stupidity.

[Borek]

I have a feeling you are doing it backwards all the time.

http://www.titrations.info/titration-calculation

You have to calculate amount of sodium carbonate in the sample from the known amount of HCl used.

[fledarmus]

Ok lets see if I am on the right track here

For my trial 1
.7893g and it took 41.12 ML of HCl to neutralize the solution

Mol Na2CO3 =.7893g

This isn't the mass of Na2CO3, it is the total amount of ash which you are trying to analyze to see what percentage is Na2CO3. You will need to go the other direction, from the .003174 moles of HCl that you calculated previously. As you also mentioned, it takes 2 moles of HCl to neutralize each mole of Na2CO3. So how many moles of Na2CO3 were in the sample? How much would that much Na2CO3 weigh? and what percentage is that of your total weight of ash?

.7893 grams of soda ash
41.12mL of HCl

mol= .7893g  X 1mol/106=.007446

This is your mistake - you've done exactly the same thing again. The 0.7893 g is NOT the weight of sodium carbonate, so you can't just divide it by the molecular weight of sodium carbonate to find the number of moles of sodium carbonate. 0.7893 g is the mass of the ASH, some percentage of which is sodium carbonate. What you are trying to do in your experiment is use a titration with HCl to find out what part of the 0.7893 g is sodium carbonate.

Think through your experiment step by step. The first thing you did was to titrate an HCl solution to find out exactly what the concentration of HCl in the solution was. You found out that your HCl solution was 0.093298 M.

Then you dissolved a 0.7893 g portion of your ash in water (or at least all of the sodium carbonate in the ash) and titrated the resulting solution with with HCl. You found that it required 41.11 mL HCl solution to neutralize all of the sodium carbonate in that sample. From this your were able to calculate that it took 0.003174 moles of HCl to react with all of the sodium carbonate.

You have also mentioned that it requires 2 moles of HCl to neutralize 1 mole of Na₂CO₃. You know how many moles of HCl you used (0.003174 moles) - with just this single observation, that you need 2 moles of HCl to neutralize 1 mole of Na₂CO₃ and that you used 0.003174 moles of HCl, how many moles of Na₂CO₃ did you neutralize?

Once you know how many moles of Na₂CO₃ you neutralized, THEN you can use the molecular weight of the sodium carbonate to find how many grams of sodium carbonate were in the ash sample, and use that number and the total mass of the ash sample to find what percentage of the ash was sodium carbonate.

[bornn]

Ok lets see if I am on the right track here

For my trial 1
.7893g and it took 41.12 ML of HCl to neutralize the solution

Mol Na2CO3 =.7893g

This isn't the mass of Na2CO3, it is the total amount of ash which you are trying to analyze to see what percentage is Na2CO3. You will need to go the other direction, from the .003174 moles of HCl that you calculated previously. As you also mentioned, it takes 2 moles of HCl to neutralize each mole of Na2CO3. So how many moles of Na2CO3 were in the sample? How much would that much Na2CO3 weigh? and what percentage is that of your total weight of ash?

.7893 grams of soda ash
41.12mL of HCl

mol= .7893g  X 1mol/106=.007446

This is your mistake - you've done exactly the same thing again. The 0.7893 g is NOT the weight of sodium carbonate, so you can't just divide it by the molecular weight of sodium carbonate to find the number of moles of sodium carbonate. 0.7893 g is the mass of the ASH, some percentage of which is sodium carbonate. What you are trying to do in your experiment is use a titration with HCl to find out what part of the 0.7893 g is sodium carbonate.

Think through your experiment step by step. The first thing you did was to titrate an HCl solution to find out exactly what the concentration of HCl in the solution was. You found out that your HCl solution was 0.093298 M.

Then you dissolved a 0.7893 g portion of your ash in water (or at least all of the sodium carbonate in the ash) and titrated the resulting solution with with HCl. You found that it required 41.11 mL HCl solution to neutralize all of the sodium carbonate in that sample. From this your were able to calculate that it took 0.003174 moles of HCl to react with all of the sodium carbonate.

You have also mentioned that it requires 2 moles of HCl to neutralize 1 mole of Na₂CO₃. You know how many moles of HCl you used (0.003174 moles) - with just this single observation, that you need 2 moles of HCl to neutralize 1 mole of Na₂CO₃ and that you used 0.003174 moles of HCl, how many moles of Na₂CO₃ did you neutralize?

Once you know how many moles of Na₂CO₃ you neutralized, THEN you can use the molecular weight of the sodium carbonate to find how many grams of sodium carbonate were in the ash sample, and use that number and the total mass of the ash sample to find what percentage of the ash was sodium carbonate.

Those moles calculations for HCl was calculated from the data I gathered from the standardization of HCl not the second part of the experiment with the unknown.I don't even know how to get the moles of HCl in the second part; I tried to but i guess I am doing it backwards?

Unless I use the average morality I found in part one for the standardization HCl and then calculate it by 0.04412 L X .092991 M = .004103 moles of HCl.
Is this correct?

Also I failed to mention that i preparation of 0.1 M HCl solution by putting *ml of concentrated HCl and then mixed with deionized water until 1 L of volume was reached. I suppose i use the avg M of HCl i found with my three trials or do i use 0.1 M HCl.

[Borek]

Have you read the page on titration calculation I linked to?

[bornn]

Have you read the page on titration calculation I linked to?

yes and that's what I got from it.

[fledarmus]

Ok lets see if I am on the right track here

For my trial 1
.7893g and it took 41.12 ML of HCl to neutralize the solution

Mol Na2CO3 =.7893g

This isn't the mass of Na2CO3, it is the total amount of ash which you are trying to analyze to see what percentage is Na2CO3. You will need to go the other direction, from the .003174 moles of HCl that you calculated previously. As you also mentioned, it takes 2 moles of HCl to neutralize each mole of Na2CO3. So how many moles of Na2CO3 were in the sample? How much would that much Na2CO3 weigh? and what percentage is that of your total weight of ash?

.7893 grams of soda ash
41.12mL of HCl

mol= .7893g  X 1mol/106=.007446

This is your mistake - you've done exactly the same thing again. The 0.7893 g is NOT the weight of sodium carbonate, so you can't just divide it by the molecular weight of sodium carbonate to find the number of moles of sodium carbonate. 0.7893 g is the mass of the ASH, some percentage of which is sodium carbonate. What you are trying to do in your experiment is use a titration with HCl to find out what part of the 0.7893 g is sodium carbonate.

Think through your experiment step by step. The first thing you did was to titrate an HCl solution to find out exactly what the concentration of HCl in the solution was. You found out that your HCl solution was 0.093298 M.

Then you dissolved a 0.7893 g portion of your ash in water (or at least all of the sodium carbonate in the ash) and titrated the resulting solution with with HCl. You found that it required 41.11 mL HCl solution to neutralize all of the sodium carbonate in that sample. From this your were able to calculate that it took 0.003174 moles of HCl to react with all of the sodium carbonate.

You have also mentioned that it requires 2 moles of HCl to neutralize 1 mole of Na₂CO₃. You know how many moles of HCl you used (0.003174 moles) - with just this single observation, that you need 2 moles of HCl to neutralize 1 mole of Na₂CO₃ and that you used 0.003174 moles of HCl, how many moles of Na₂CO₃ did you neutralize?

Once you know how many moles of Na₂CO₃ you neutralized, THEN you can use the molecular weight of the sodium carbonate to find how many grams of sodium carbonate were in the ash sample, and use that number and the total mass of the ash sample to find what percentage of the ash was sodium carbonate.

Those moles calculations for HCl was calculated from the data I gathered from the standardization of HCl not the second part of the experiment with the unknown.I don't even know how to get the moles of HCl in the second part; I tried to but i guess I am doing it backwards?

Unless I use the average morality I found in part one for the standardization HCl and then calculate it by 0.04412 L X .092991 M = .004103 moles of HCl.
Is this correct?

Yes, this is correct - I'm sorry, I thought those were the numbers you were reporting in your second post. (.092991 M or .93298 M?)

Also I failed to mention that i preparation of 0.1 M HCl solution by putting *ml of concentrated HCl and then mixed with deionized water until 1 L of volume was reached. I suppose i use the avg M of HCl i found with my three trials or do i use 0.1 M HCl.

Yes, you use the average of your 3 trials. Your titration should be more accurate than your preparation, due to variations of the concentration of HCl in commercial concentrated HCl solutions

[bornn]

ok I think I got it finally.

0.04412 L X .092991 M =.004103 moles of HCl

HCl reacts 2:1 ratio with Na2CO3 therefore,

.004103 X 1/2 = .002052 mol Na2CO3

molecular weight of Na2CO3=105.99g/mol

.002052 mol X 105.99 g/mol=.217491g

.217491g/.7893g X 100% =27.5%

Also does anyone know if this is close to the accepted value of Na2CO3 in soda ash cant seem to find info on that.

Also thank you everyone for all the help I really do appreciate it.