[the_guitarist]

   To find the concentration of my Na₂CO₃ solution I used the following formulae:
Mass=Moles x Relative Molecular Mass
Moles=Concentration x Volume

Firstly, I worked out that Na₂CO₃ has a relative molecular mass (RMM)  of 106:
2Na=23 x2=46
C=12
3O=16 x 3=48
46+12+48=106

After this I calculated the number of moles in the solution by diving 2.67g by the RMM of sodium carbonate, 106.
2.67
----- =0.025mol.
106

Concentration=Moles/volume.
= 0.025/0.25dm=0.1moldm-3

The equation:
Na₂CO₃(aq)+H₂SO₄(aq)---->H₂O(l)+Na₂SO₄(aq)+CO₂(g)

Because the molar ratio in this equation is 1:1 the volume ratio will be 25:20, meaning that the calculation to work out the concentration of the H₂SO₄ will be this:
Moles=Concentration x Volume
1:1 moles = 20ml:25ml

H2SO4= 0.025
              ------- =0.125 moldm^-3
                0.2
The solution of the H₂SO₄ is 0.125 moldm^-3

If anyone could take a quick look over that and check my calculations I would greatly appreciate it.

[Borek]

If you will start explaining what you are calculating, it may help a little bit.

[the_guitarist]

I'm calculating the concentration of my sulphuric acid from a titration.

[Borek]

Listing titrant volume won't hurt.

[the_guitarist]

It took 20ml of sulphuric acid to reach the end point in 25ml of sodium carbonate

[Guitarmaniac86]

Edited in post below due to errors.

[Guitarmaniac86]

RMM of Sodium Carbonate = 2.67 / 106 = 0.025 moles.

That goes in 250 cm volumetric filled to line so the concentration is: (0.025 x 1000)/ 250 = 0.1 mol dm^3

In 25 cm^3 you have (25 x 0.1)/ 1000 = 2.5 x 10^-3 moles.

Reaction is 1:1

H2SO4 Conc = (2.5x10^-3 x 1000) / 20 = 0.125 mol dm^-3

Yay! I got it... I think.

God damn vodka coursing through me.

[Borek]

0.125 seems OK. In future please list your question with description of what you did - something like "Took 2.67 g of Na2CO3, dissolved in 250 mL of water, took 25 mL of that and titrated with 20 mL of H2SO4, what is acid concentration". Otherwise we have to guess what and why and how you are doing and it is very difficult to check/help.

[Guitarmaniac86]

Tim (I mean the_guitarist) there is a method you should learn when doing these kind of calculations.

I learnt this step wise and the more calculations you do, the easier it becomes until its second nature. You need to sit down and devise a method of learning the steps to this calculation.

Step one: The stock solution

If you are asked to make a certain concentration of a stock solution, you need to know

-The volume that the solution is going to be made into
-The molarity of the solution (concentration) you want to make
-The number of moles required to make that concentration at that volume.

The formula that you need to know backwards is Moles = (Concentration x Volume) / 1000.  You need to know that and know how to manipulate it.

To find concentration you re arrange that equation to: Concentration = (Moles x 1000) / Volume.

To find the volume you re arrange the original equation to: Volume = (Moles x 1000) / Concentration.

To avoid confusion, keep everything in cm3 or ml. Dont convert to dm^-3 unless you are sure you can manipulate the numbers. Check using both methods (one keeping it in ml/cm3 and the one where the volume is converted to dm^3) to see if they both give the same answer.

Worked Example For Stock Solution Preparation

You are asked to make a solution of 0.2 Mol dm^-3 of NaOH in a 250 cm3 volumetric flask.

The first thing you need to do is work out how many moles of NaOH you need to have in that solution. So use Moles = (Conc x Volume)/1000

Moles NaOH = (0.2 x 250)/1000

Moles NaOH = 0.05 Moles

Now you need to find the mass of NaOH that is 0.05 moles. For this use Moles = Mass/RMM when re arranged to find mass:

Mass = Moles x RMM.

The RMM of NaOH is: 40 therefore the mass of NaOH needed is: 0.05 x 40 = 2 grams

Summary

If you are asked to make a stock solution of known concentration you need to know:

-The volume of the solution you will be making
-The number of moles required to make that concentration using Moles = (C x V) / 1000
-From that you need to know the mass of the substance that will be used to make the solution and you use Mass = Moles x RMM

Step Two: The Titration

Say we take 25 cm3 of NaOH from the stock solution and place it in a conical flask with a suitable indicator and react it with an unknown concentration of HCl. We will need to know how many moles of the NaOH is present in the 25cm3. Clearly there will be 10x less moles of NaOH in the solution so we use:

Moles = (C x V)/ 1000 = (0.2 x 25) / 1000 = 0.005 x 10^-3 Moles in the 25 cm3 solution that was taken from the stock.

When the titration is performed, 22.3 cm3 of unknown HCl reacted with the NaOH. Write the reaction out to find the ratio of the reaction:

NaOH + HCl --> NaCl + H2O

Clearly the reaction is 1:1

From this we can find the concentration because if the reaction is 1:1 0.005 moles of NaOH will have reacted with 0.005 moles of HCl. We also know the volume of HCl that reacted. It was 22.3cm3.

We can now use Concentration = (Moles x 1000) / Volume = (0.005 x 1000) / 22.3

For HCl therefore the Concentration was: 0.224 mol dm^-3

Summary

-Find the number of moles in the portion of the known solution taken from the stock solution
-Find the volume of the unknown that reacts with the known solution
-Write the reaction out to find the mole ration
-Use the mole ratio and volume to work out the concentration using Concentration = (Moles x 1000) / Volume.

Tim (I mean the_guitarist).. If you follow these steps and keep using these steps you cant go wrong with titrations. No matter what titration you do, you will only use two formula: Moles = (C x V) / 1000 and Moles = Mass / RMM

I hope this helps you with every question you will ever face with titrations.