- Created by: abi elizabeth
- Created on: 17-05-19 14:04
Ideal gases are gases that have:
- Molecules in random motion
- No intermolecular forces present
- Elastic collisions with no exchange of kinetic energy
- particles of zero volume
Real gases are unlikely to follow these conditions exactly but do get close at high temps and low pressures
the volume of gas is directly linked to:
- Avagadro's law: Higher moles = Higher volume (V is directly proportional to n)
- Charles' Law: Higher temps = Higher volume (V is directly proportional to T)
- Boyle's Law: Higher pressure = Higher volume (V is directly proportional to p)
All of those laws an be combined to gain the ideal gas equation:
PV = nRT
P = pressure (pa)
V= volume (m3)
n= Number of moles
R= Gas constant (8.314)
T= Temperature (K)
Try to work this out:
What volume would 12.8g of oxygen gas have at 15 degrees celcius and a pressure of 90Kpa?
answer: V= 0.0106 m3
Atom economy and percentage yield
Atom economy is the total number of atoms in the required product divided by the total number of atoms in the products
Percentage yield is worked out by the equation
Percentage yield = mass from practical of desired product (actual mass) / Theoretical mass
Theoretical vs actual mass
- incomplete reaction
- reversible reaction (only reversible in reaction)
- Another unexpected side product
- transferring between vessels
Titrations are used to find out the concentration of an acid or alkali solution.
A standard solution has a known concentration:
e.g. Make 250cm3 of a 2.00 mol dm-3 solution of sodium hydroxide
1.) First work out how many moles of sodium hydroxide you need using this formula moles= concentration x volume = 2.00 mol dm-3 x 0.250dm3 = 0.500 moles
2.) Now work out how many grams of sodium hydroxide you need using this formula; mass = moles x mr = 0.500 x 40.0 = 20.0g
3.) Place a weighing boat on a digital balance and weigh out this mass of a solid. Tip it into a beaker. Now re-weigh the boat and subtract the mass of the boat from the mass of the boat and solid together to find the precise mass of the solid used.
4.) Add distilled water to the beaker and stir until all the sodium hydroxide has dissolved
5.) Tip the solution into the volumetric flask - making sure that it is the right volume that you are making (250cm3 in this case). use a funnel to make sure it all goes in
6.) Rinse the beaker and stirring rod with distilled water and add that to the flask too. This makes sure that there's no solute clinging to the beaker or rod.
7.) Now top up the flask up to the correct volume with more distilled water. Make sure the bottom of the meniscus reaches the line. When you get too close to the line, add water dropwise, as if you go over the line you will have to start all over again.
8.) stopper the flask and invert the flask a few times to ensure the mixture is mixed
Purpose of a titration
A titration allows you to find out exactly how much acid is needed to neutralise a measured quantity of alkali (or the other way round). You can use this data to work out the concentration of the alkali. Start of by using a pipette to measure out a set volume of the solution that you want to know the concentration of. put it in a flask. Add a few drops of an appropriate indicatior to the flask. Then fill a burette with a standard solution of the acid. Use a funnel to pour the acid into the burette to ensure that you avoid spalshing acid. You are now ready to titrate..
Titrations need to be done really accurately
1.) Firstly, do a rough titration to get an idea of where the end point is. Add the acid to the alkali using the burette, giving the flask a regular swirl.
2.) Now do an accurate titration. Take an initial reading to see exactly how much acid is in the burette. Then run the acid in to within 2cm3 of the end point. When you get to this stage, add it dropwise - if you don't notice exactly when the colour changes, you'll overshoot and your result won't be accurate.
3.)Work out the amount of acid used to neutralise the alkali. This is just the final reading minus the initial reading. This volume is known as the titre.
4.) Repeat the titration a few times, until you have at least three results that are concordant (very similar)
5.) Use the results from each repeat to calculate the mean volume of acid used. Remember to leave out any anomalous results when calculating you mean- they can distort your answer.
Indicators show you when the reactions just finished:
In titrations, indicators that change colour quickly over a very small pH range are used so you know exactly when the reaction has ended.
The main two indicators for acid/alkali reactions are:
1.) Methyl orange - This is red in acid and yellow in alkali
2.) Phenolphthalein- this is colourless in acid and pink in alkali
It's a good idea to stand your flask on a white tile - to make it easier to see exactly when the end point is.
e.g. In a titration experiment, 25.0cm3 of 0.500 mol dm-3 HCl neutralised 35.0cm3 of NaOH solution. Calculate the concentration of the sodium hydroxide solution in mol dm-3.
1.) First write a balanced equation and decide what you want to know and what you need to know:
HCl + NaOH -> NaCl + H2O
0.500 mol dm-3 ?
2.) Now work out how many moles of HCl you have:
Number of moles of HCl = concentration x volume (dm3) = 0.500 x (25.0/1000) = 0.0125 moles
3.) From the equation, you know that 1 mole of HCl neutralises 1 mole of NaOH. So 0.0125 moles of HCl must neutralise 0.0125 moles of NaOH: conc of NaoH = 0.0125/ (35/1000) = 0.357
Calculating volumes for reactions
e.g. 20.4 cm3 of a 0.500 mol dm-3 solution of sodium carbonate reacts with 1.50 mol dm-3 nitric acid. Calculate the volume of nitric acid required to neutralise the sodium carbonate.
1.) First write a balanced equation for the reaction and decide what you know and need to know:
Na2CO3 + 2HNO3 --> 2NaNO3 + H2O + CO2
20.4 cm3 ?
0.500 mol dm-3 1.50 mol dm-3
2.) Now work out how many moles of Na2CO3 you have:
number of moles = conc x volume (dm3) = 0.500 x (20.4/1000) = 0.0102 moles
3.) 1 mole of Na2CO3 neutralises 2 moles of HNO3 - moles of HNO3 = 0.0204 moles
volume of HNO3 = moles/conc = 0.0204/1.50 = 0.0136 dm3