[wicked1977]

Hello, first time posting here.

My teacher proposed a scenario in which we find the molarity of HCl. Here are the guidelines:

1) You have been hired by Titrations-R-Us to determine the molarity of HCl present in industrial waste samples (usually in the range of 0.200 to 0.300M. Results to 3 sig figs.

2) Normal procedure is by titration using a standard solution of 0.100M NaOH and a phenolphthalein indicator.

3) All lab equipment are present except for calibrated glassware cannot be used (pretty much anything that is precise like burettes, graduated cylinders, volumetric flasks, etc.) We can use beakers and such since it is not accurate.

4) Explain how you might perform this analysis & maintain the necessary 3 sig fig result. Provide an explanation with a hypothetical numerical example.

I was searching through the interwebs and found a procedure (reference link: http://www.hschem.org/Laboratory/WoodrowWilsonFoundLabs/acidbasetitrations.html)

1) Fill one "buret" (plastic bottle) with the 5.00% NaOH solution. Determine the mass of the filled "buret" and contents and record.

2) Fill the other "buret" with the HCl stock solution. Determine the mass and record.

3) Transfer 10-20 g of HCl solution (about 1/3 of the contents of a 2-oz "buret") to a flask. Add 3 drops phenolphthalein solution. Gradually add the NaOH solution, with swirling, until the pink phenolphthalein color begins to persist. When near the endpoint, add the NaOH solution dropwise until one drop causes a pink color that persists for 30 seconds. If the endpoint is over-run, more acid may be added and the titration continued.

4) Remass both burets.

5) Repeat steps 1-4 two more times.

6) From the mass of NaOH solution used in each trial, calculate the mass of HCl in the sample.

7) Calculate the percent of HCl in the solution used in each trial.

Calculate the average percent of HCl in the solution.

Following the procedure above, I used three 100mL beakers, one with the HCl and the other with the NaOH. Following the procedure, I made a hypothetical data to get a hypothetical number.

Beaker Mass: 50.000g
Mass of Beaker + NaOH solution: 70.000g
Mass of Beaker + HCl solution: 70.000g
Mass of Beaker + NaOH solution after titrating: 60.000g
Mass of Beaker + HCl solution: 60.000g

So I have 10g of NaOH used, but I do not know how to use this number to find mass of HCl in the sample. I thought of doing stoich such as:

NaOH + HCl:rarrow:NaCl + H₂O

Converting 10g of NaOH to 10mL (not sure if you can do this haha)

0.01L NaOH x (0.500 mol NaOH/1L NaOH) x (1 mol HCl/1 mol NaOH) = 0.005 mol HCl

Density of HCl: 1.49

1.49 g/l = 10g/x (where x is L
x= 6.711 liter

0.005mol HCl/6.711L = 7.45x10^-4 completely off the range!

Any help will be appreciated.

[mikasaur]

The steps you got from the internet look pretty good. Your hypothetical experiment has a lot of problems, though.

One of your problems is assuming 10 g of NaOH solution is 10 mL. You can't do that, unfortunately.

Are you given the density of 0.100M NaOH? Are you allowed to use that?

[Borek]

Density of HCl: 1.49

That's (in g/L) the density of gaseous HCl, you are dealing with the aqueous solution, so this number is three orders of magnitude off.

As a first approximation you can assume all solutions involved to be diluted enough so that their densities are that of a pure water. As mikasaur pointed out, that can be a source of errors. However, you are asked for a limited accuracy only. Google for density tables and check, if for the concentrations given, the approximation holds within 3 sig figs limit.

[wicked1977]

I have been told that if I find a flaw in my method for solving the scenario, I can get extra points for it. Thus, yea I can assume that because all the reactants are in aqueous solution, their densities are near pure water: 1g/L (so this would be a flaw but I do not know excatly why it is though because isn't density the same for every element/compound?) Regarding density of NaOH, yea I can use that but would I use the actual density or assume the density for both reactants are 1g/L? I'll try these later.

[wicked1977]

Density of HCl: 1.49

That's (in g/L) the density of gaseous HCl, you are dealing with the aqueous solution, so this number is three orders of magnitude off.

As a first approximation you can assume all solutions involved to be diluted enough so that their densities are that of a pure water. As mikasaur pointed out, that can be a source of errors. However, you are asked for a limited accuracy only. Google for density tables and check, if for the concentrations given, the approximation holds within 3 sig figs limit.

I tried doing the calculations but I cannot logically think of a way to get the percent of HCl (assumed that NaOH and HCl have densities of 1g/mL). Any guidelines would be greatly appreciated!

[Borek]

I tried doing the calculations but I cannot logically think of a way to get the percent of HCl (assumed that NaOH and HCl have densities of 1g/mL). Any guidelines would be greatly appreciated!

Hard to say where your problem is, but you probably sinfusing several tghings.

First of all - each time you refer to NaOH or HCl, you say something like "I have 10 g of NaOH". You don't. You have 10 g of a solution of NaOH. That's not the same.

Also note:

Beaker Mass: 50.000g
Mass of Beaker + NaOH solution: 70.000g
Mass of Beaker + HCl solution: 70.000g
Mass of Beaker + NaOH solution after titrating: 60.000g
Mass of Beaker + HCl solution: 60.000g

This can't be right, unless I am misreading what you wrote. Your procedure was to transfer NaOH solution from the NaOH beaker to the HCl beaker, yes? If so, mass removed from the NaOH beaker went to the HCl beaker, so the final mass of HCl beaker should be 70+10 g, not 60 g.

What was mass of the HCl solution in the HCl beaker?

You know number of moles of HCl present in the beaker, how many grams is it?

Look at the last two numbers. Are they enough to calculate HCl %?

[wicked1977]

First, I add 10g of HCl solution to the third beaker. Second, I gradually add NaOH solution to the third beaker until I reach an endpoint and thus assuming that 10g of the NaOH solution is only needed to reach it. I should really specified, sorry.

Based on what you said I believe it is enough to calculate HCl %? The main goal is to find the concentration of HCl anyways so this is what I did.

So there are 10g of HCl solution and 10g of NaOH solution used, so 10g of HCl solution and 10g of NaOH are left.

10g NaOH x 1mol NaOH/39.997g NaOH x 1mol HCl/1mol NaOH x 36.46g HCl/1mol HCl = 9.12g HCl
9.12g/10g x 100% = 91.2%

Assume 100g of solution
91.2g/36.46g/mol = 2.49mol HCl

1g/mL = 100g/x mL
x=100mL

2.49mol HCl/0.1L = 24.9M HCl
Correct or no?

I think the molarity in this hypothetical example is big because we assumed density and random grams of solutions.

[Borek]

10g NaOH x 1mol NaOH/39.997g NaOH

You see, that's what I was talking about. 10 g of NaOH is not the same thing as 10 g of NaOH solution. You don't have 10 g of NaOH! Convert solution mass to the solution volume and calculate number of moles of NaOH from the solution volume and NaOH concentration.

[wicked1977]

So pretty much instead of 10g of NaOH solution its 10mL of NaOH solution? So I do the same thing except
10mL x 1L/1000mL x 0.500mol/L NaOH = 0.0005 mol and I follow through with the rest of the procedure

[mikasaur]

Yes, except I think your original prompt has 0.100 M NaOH, not 0.500 M.