Standardization of HCl using NaOH and Na2CO3

[Chu]

I need help filling in the values for [HCl] (M) - USED (for both parts) and [HCl} (M) - STOCK (for part 1)

 

More specifically I need to know the equations and the inputs (like the area in the f_x writing space) that would result in my average, standard deviation, and RSD % resulting in numbers that would match the blue boxes.

[hypervalent_iodine]

If this is homework, you need to make some attempt at the questions. What equations have you learned so far? 

[Chu]

I have tried but every attempt leaves me with incorrect answers so I just left those blank for the purpose of this forum. Everything in the gray is what I have/know how to fill out. The teachers aren't telling us students anything and are kind of just throwing us out here with no idea what to do really. 

[hypervalent_iodine]

Okay, so what are your attempts? What equations have you been using?

Could you also clarify for me the first table. Are you adding the HCl to the Na2CO3 and then the NaOH to that solution once it finished reacting? What does he [HCl] used refer to exactly? Used in which bit?

[Chu]

For the first table NaOH is being titrated from the buret into a erlenmeyer flask containing the Na2CO3,  DI water, and 50 mL HCl (which is of course the flask is boiled first). Then it is titrated to the endpoint. The results are as follows in the excel image. 

Calculations I have tried (this is all they have taught us to use):

g of Na2CO3 to find g of HCl and then attempting to find the M of HCL used and stock, but I just get stuck after g HCl

(stock NaOH * Vol NaOH used)/(****)/(vol flask/vol pipet)             (found this one online and I feel like its the closest but don'd know how to apply it)

           ((2.052M * 25.75 mL)/****)/(249.74/24.996)

I think what I am really getting stuck on is the dilution part between the stock solutions and the 10x dilution of 25 mL into a 250mL vol flask, as I have never done this before and now I am running out of time to do the assignment. It is due tomorrow (Sunday) at 5 PST.

I forgot to mention that I have tried other equations, I just forgot to write them down as I did them in excel and quickly found out that they were wrong. And as for table 2 it is just a standard direct titration between NaOH (buret) and HCl (flask).

[hypervalent_iodine]

Your first calculation isn't going to work. The entire point is to standardise the HCl, so you're supposed to work it out based on the NaOH titration. You have two things going on here; a reaction between the Na2CO3 and HCl, and a reaction between whatever HCl didn't react and the NaOH. The idea is that you are supposed to react all of the Na2CO3 with only some of the HCl (ie. there is excess HCl and limited Na2CO3 available). This leaves a certain number of moles if HCl left available to react, so you use the NaOH for this part. The number of moles of HCl is therefore a combination of the moles used in both reactions, not just the one. Honestly, this seems like a pretty silly way to standardise anything (why not just titration with NaOH?), but that's neither here nor there I guess. 

The C1V1=C2V2 is not a great calculation in this instance because you have two things going on. In a normal titration it would work fine, but not this one. 

Forget about the dilution. You didn't dilute any of the stocks prior to combining them, so it doesn't matter. All you need to work with is moles and the original volumes that you added.

 

 

 

[Chu]

I don't think I am following you. All I can think of is using a density equation which I know is incorrect. I don't see how the mL of HCl is suppose to convert into an equation with the Na2CO3 that would result in the molarity of HCl used.

[hypervalent_iodine]

Can you give an equation for the reaction between the two?

As in A + B --> C + D not a mathematical equation

[Chu]

You mean a balanced chemical equation like this?

Na2CO3 + 2HCl → 2NaCl + CO2 + H2O

[hypervalent_iodine]

Yep!

The short story is that you need to know the total number of moles of HCl you have in the flask. You perform two reactions with the same volume of HCl, so the total number of moles is equal to the number of moles used in the first reaction plus the number of moles used in the second.

The two reactions:

1. In your experiment the first thing that you do is the reaction you wrote out above. You mix HCl with Na2CO3, and you make NaCl, CO2, and water. You assume that you used up all of the Na2CO3 in the initial reaction, and that there was an excess of HCl (do you know what I mean by this?). The number of moles of HCl used up will be equal to half the number of moles of Na2CO3 that reacted, based on the stoicheometric equivalents in your chemical equation. 

2. You then have some HCl left over that wasn't able to react with the Na2CO3 (otherwise you wouldn't have titrated with NaOH), and you react this with NaOH until all of the HCl is gone. Do you know the chemical equation for this reaction?

To help illustrate what I am getting at, consider a case similar to yours where you have 1 g of Na2CO3 in a flask. This is equal to .0095 moles of Na2CO3. To this we add 50 mL of a 1M HCl solution. In that 50 mL there is 0.05 moles of HCl. Based on the stoicheomtry of the equation, if all of the Na2CO3  were to react, it would need 0.019 moles of HCl. This is less than the amount of HCl that we have available, so all of the Na2CO3 will react and 0.031 moles of HCl will be left unreacted. If we then want to check how many moles of HCl is left, we could titrated against NaOH to react what was remaining. Let's say we do the titration with 1M NaOH. Given that there is 0.031 moles of HCl left, if you did the titration would find that you would need on average, 31 mL of 1M NaOH to get to the end point and react with all the HCl available. Working backwards, if we didn't know what the concentration of HCl was, we could figure out that in the first reaction with Na2CO3 we needed 0.019 moles, and in the second reaction with NaOH we needed 0.031 moles. Thus the total number of moles in the 50 mL of HCl we added is 0.031 + 0.019 moles, which is 0.05 moles. The concentration is therefore 0.05 moles / 0.05 L, which of course equals 1M. 

Does that make sense?

[Chu]

I tried what you said, but I think I am missing something. For trial 1 these were my results:

0.1298g Na2CO3 yielded 0.0024493 mol HCl used the first reaction. Then converted I 25.75 mL NaOH to 0.052839 mol NaOH given the M of 2.052 from the stock.

Since I didn't know the concentration of HCl I added 0.0024493 + 0.052839 = 0.0545332 mol HCl 

I then divided 0.0545332 by 0.05L (from the 50 mL HCl used and ended up with 1.091M HCl.

This yield is too small of a number to fall within one standard deviation of the number in the blue box in the picture and therefore cannot be right. I even tried to solve for the other 2 trials and found the average to be 1.112.

[hypervalent_iodine]

Your math seems fine. There's nothing you can do about it if that's what the results are giving you. What did you use to determine the end point?

[Chu]

I got the endpoint from the mL of total NaOH used. G10-G12 from the picture. 

[hypervalent_iodine]

Did you not do this experiment yourself? I meant how did you determine it experimentally - ie. when did you decide you had added enough NaOH?

[Chu]

I haven't done it yet. Everything but the gray comes pre filled and we are expected to learn it and turn it in before we can do the lab. I pretty much gave you everything they gave me. The only thing I did was subtract initial volume from final.

[hypervalent_iodine]

Okay, so here's my suspicion. They made a typo. If you use 1.298 g of Na2CO3 instead of 0.1298 g, you get the correct answer. My advice is to contact your course coordinator and check with them if this is the case.

[Chu]

Okay, I will do that ASAP!

Thank you for all your help it's really appreciated.

[Chu]

I thought that you'd like to know that I figured some of it out!

In order to find the [HCl] (M) - STOCK I had to use this equation for trial 1:

(((0.1298 g Na2CO3*10)*(1 mol/105.9888 g)*(2 mol HCl/1 mol Na2CO3)+(25.75 mL NaOH used*(1 L/1000 mL)*(2.052 M NaOH STOCK/ 1 L)))/0.05 L

[hypervalent_iodine]

That looks good to me. Well done!