Calculations Involving Titrations

Created on Monday, May 2, 2011.

Filed under
Life Skills.

You go to the supermarket and buy a 3L bottle of bleach for $3.40. Scandalised by the realisation that most of this is probably water, you resolve to go home and calculate exactly how much of the active ingredient (molecular Chlorine) you have actually paid for.

Let’s begin with my favourite Chemistry joke:

Argon walks into a bar, and the bartender yells, “Get out! We don’t serve noble gases here!”

Argon doesn’t react.

Here is a complete stoichiometry calculation, from initial titration to final %w/v value and weight-to-cost price. I could have put this on the university message boards, but they probably delete the board contents after every semester, and this kind of thing will be useful for posterity. Also, the uni’s HTML creator is broken, and the one I wrote is better anyway.

The stoichiometry problem I’ll be explaining is as follows:

Back at your mom’s basement, you take 3 mL aliquots of this bleach sample and, having diluted it and mixed it with excess acid and an iodine indicator under adequate ventilation, you titrate it with a 0.1010±0.0008M solution of Na_{2}S_{2}O_{3}. You find that, on average, you need to add 15.49 mL of this titrant in order to reach the endpoint.

Calculate the %w/v of Sodium Hypochlorite (NaOCl) and the %w/v of molecular Chlorine (Cl_{2}) which were actually present in the bleach to see if they measure up to the manufacturer’s promises, and then calculate how much you paid for each gram of Cl_{2} so that you can be more informed next time.

- Bottle volume: 3000 mL = 3L
- Price of bottle: $3.40 = 340 cents
- Aliquot volume: 3.00±0.05 mL = 0.003±0.00005 L

- Molar concentration: 0.1010±0.0008 mol/L
- Average titre volume: 15.49±0.06 mL = 0.01549±0.00006 L

- Molar mass of NaOCl: ~74.44 g/mol
- Molar mass of Cl
_{2}: ~70.90 g/mol - Stoichiometric ratio*: 1 mol of titrant : 0.5 mol of iodine indicator
- Iodine-to-NaOCl ratio*: 1:1 ratio

**Find the chemical amount of titrant used in the reaction, using n=cV.**

n=cV -> n=0.1010*mol/L*× 0.01549*L*= 0.00156449 mol Na_{2}S_{2}O_{3}.**Use the stoichiometric ratio to find the chemical amount of indicator that participated in the reaction, by multiplying the chemical amount of titrant by 0.5.**

Bringing a reaction to its endpoint means that you have added enough titrant to the analyte to make the indicator completely react and change (either through direct reaction, as in this case, or in reaction to the analyte which has speciated etc). The stoichiometric ratio (1:0.5 of titrant to indicator) tells us that if you have 0.5 mol of indicator, you need 1 mol of titrant to reach endpoint.

0.00156449*mol*× 0.5 = 0.000782245 mol of I_{2}(from indicator).**Find the chemical amount of NaOCl by multiplying with its ratio to indicator.**

Since iodine and NaOCl react equally with each other (a 1:1 ratio), there is the same amount of NaOCl as there is I_{2}.

0.000782245*mol*× 1 = 0.000782245 mol of NaOCl**Now that you know the chemical amount of NaOCl, and the original volume of the aliquot, you can calculate the concentration of NaOCl in your bleach sample.**

The concentration of your 3 mL aliquot will be the same as the concentration in the 3 L bottle of bleach you took it from. No further wizardry is necessary.

n=cV -> c=n ÷ V -> 0.000782245*mol*÷ 0.003 L = 0.26074833 mol/L of NaOCl.

**Divide the Molar concentration of NaOCl by 10.**

In the last section we found the Molar concentration of NaOCl. What is Molar concentration? It is simply moles per Litre. This means that dividing the concentration by 10 will give us moles per 100 mL. Kapow!

0.26074833 mol/L ÷ 10 = 0.026074833 mol of NaOCl per 0.100 L**Use m=nM to find the mass of NaOCl present in 100 mL of bleach.**

The Molar mass (M) of NaOCl is ~74.44 g/mol.

0.026074833 mol/0.100 L × 74.44 = 1.941010569 g of NaOCl per 100 mL

∴ the bleach contains 1.94%w/v of NaOCl.

**Divide the Molar concentration of NaOCl by 10.**

It turns out that in the presence of lots of acid, NaOCl releases molecular chlorine (Cl_{2}) at a 1:1 ratio, so the chemical amount of Cl_{2}is the same amount as NaOCl. This means we can apply the same divide-by-ten trick.

0.26074833 mol/L ÷ 10 = 0.026074833 mol of Cl_{2}per 0.100 L**Use m=nM to find the mass of Cl**_{2}present in 100 mL of bleach.

The Molar mass (M) of Cl_{2}is ~70.90 g/mol.

0.026074833 mol/0.100 L × 70.90 = 1.84870566 g of Cl_{2}per 100 mL

∴ the bleach contains 1.85%w/v of Cl_{2}.

**Multiply the %w/v of Cl**_{2}by 30 to get total mass of Cl_{2}in 3 L bottle.

Since the %w/v described g/100 mL, we multiply it by 30 to get g/3000 mL.

1.85 × 30 = 55.5g of Cl_{2}per 3000 mL**Divide the total cost by the total mass of Cl**_{2}to get the cost of Cl_{2}in cents/g.

340 cents ÷ 55.5g = 6.126126126 cents/g = 6.13 cents/g of Cl_{2}.

†: You may be wondering why, when I was decribing the preparation of the analyte, I didn’t outline how much the sample was diluted, how much indicator was added, and so on. It’s because it doesn’t really matter, and doesn’t factor into the final calculations that we do (which is what I am concentrating on explaining).

*: You may also be wondering how I arrived at the stoichiometric ratio and the Iodine-to-NaOCl ratio. To be honest, it was given to me, and I don’t quite understand how it was derived. I *know* that you get it by writing a net ionic equation wherein you balance the reactants and products, and you can read the actual equations and explanations here, but finding what products are made by what reactants is a bit beyond my expository capabilities right now. Just trust my numbers.