2NaOH(aq) + H2SO4(aq) --> Na2SO4(aq) + 2H2O(l)
Method (a): by measurement of the gas
Apparatus:
-the quantities of chemicals you would use and calculations to show how you decided these quantities.
- Set up apparatus as shown below.
- Weigh strontium metal
- Place 889
- Record the final volume of gas produced when reaction has occurred thoroughly (when no more bubbles are being produced).
Cant go over 100cm, gas syringe only holds 100 cm
60cm h2 gas -- 24 litres
Moles of gas
Moles of strontium
Method (b): by titration of the strontium hydroxide produced
Apparatus:
-Detailed practical methods, including a labelled diagram which give full instructions about how to obtain accurate results.
-a calculation to show how the relative atomic mass of strontium would be calculated from the titration (b) 0.1 mol/0.01 mol
-
Set up apparatus as shown below.
Accuracy:
Plan and describe how you would carry out both methods accurately.
full instructions about how to obtain accurate results.
Repeat readings, tests, maintain room temperature
Hazard & Safety Precautions:
Bibliography:
Bernadette Walsh
April 2008
Planning Exercise
Introduction:
The aim of this experiment is to demonstrate that sulphuric acid, H2SO4 is dibasic.
Scientific Knowledge:
Sulphuric acid, H2SO4, is a dibasic (diprotic) acid. This means that one mole of sulphuric acid can give off two moles of hydrogen ions, H+, when It reacts in aqueous solution.
H2SO4 can be proven to be dibasic/diprotic with two methods, by gas collection and by titration. These methods will be explained in the next section.
Methods:
Method (a): by measurement of the gas:
1. Set up apparatus as shown below:
Calculations for gas collection:
Use Excess Mg. 0.35 100 cm cubed big
Due to the fact that only 100cm3 gas syringes are available, the maximum amount of gas that can be produced in this experiment will be 0.042 moles of C02 (as 100/2400 = 0.42). For one mole of H2S04 the maxiumum volume is 0.0042 x 1000/10 = 4.2 cm3. to ensure that I do not go over this limit I will only use 3.5cm3 of sulphuric acid. To ensure that all of the H2SO4 reacts, I must use excess MgCO3 the minimum MgCO3 I can use is
Mass if MgCO3 = 0.0042 x 84 = 0.35g of MgCO3. To ensure that I am in excess you could use 0.5 of MgCO3.
MgC03 + MgSO4 + C02 + H20H2SO4(aq)
Method (b): by titration:
The reaction of sulphuric acid with sodium hydroxide solution forms sodium sulphate solution, in which both of the acidic hydrogen’s react with hydroxide ions. This reaction is as follows:
2NaOH(aq) + H2SO4(aq) --> Na2S04(aq) + 2H20
1. Set up apparatus as shown below:
2. Using a pipette/burrette and volumetric flask, dilute the H2SO4 accuratly. To make 0.1 moldm−3, mix
50 cm3 burrettes
Because H2SO4 is dibasic, I need to use twice the concentration of Na0H (0.2 moldm−3).
Calculations for titration:
2NaOH(aq) + H2SO4(aq) --> Na2S04(aq) + 2H20
Appx 10 moldm−3 H2SO4
Dilution 5.0cm3 H2SO4 in 1000 cm3 of water gives 0.05 moldm−3
5.0/1000 x 10 = 0.05 moldm−3
Assume 25cm3 of H2SO4 is required to neutralise 26cm3 of 0.1mol Na0H for neutralisation.
Accuracy:
Repeat readings, tests, maintain room temperature...
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Yahoo:
Sulphuric Acid = Dibasic
for titration(dibasic) react H2SO4 with a base sodium hydroxide and so the acid will be completely neutralized by 2 moles of those bases but for a monobasic acid(e.g. HCl) only 1 mole will be needed which will prove that H2SO4 is dibasic.
for gas collection first react a carbonate(MgCO3) with a monobasic acid(HCl)and the ratio of acid used to CO2 produced with be 2:1 but for a diprotic acid it will be 1:1. So i will only need half the amount of H2SO4 compared to HCl to produce the same amount of CO2 which will again show that H2SO4 is dibasic...
Prove that sulphuric acid is dibasic using the collection of a gas?
Based on the principle that H2SO4 + X --> XSO4 + H2
So, X could be magnesium ... that would react reasonably steadily.
So mix, say, 10 cm3 of 1M sulphuric acid with 0.3 g (an excess) of magnesium and show, by collecting the gas evolved, that about 240 cm3 of hydrogen have been produced. i.e. 0.01 mol of the acid gives 0.01 mol of hydrogen.
You may need to scale the experiment up/down to suit available apparatus. Accuracy isn't hugely important (though you should be able to explain errors) as you are only trying to show the nearest whole-number ratio.
gas collection?
Assuming you mean it dissociates into two H+ ions...
-Do a neutralisation titration (using methyl orange or an equivalent as an indicator) of an equal amount (eg: 25ml) of 1M HNO3; 1M HCl and 1M H2SO4 (separately of course), using 1M NaOH.
-Record the volumes of NaOH needed to neutralise each.
-As they are strong acids the monoprotic acids (hydrochloric and nitric) should be within a reasonable margin for error (5%-ish), but roughly the same, showing that they have an equal proportion of H+ ions to be neutralised.
H+ + OH- > H2O
-However the diprotic sulphuric acid (yes I'm English) should react with twice as much NaOH, showing it has twice as many H+ ions.
2H+ + 2OH- > 2H2O
(^needs twice as many OH ions see)
H2SO4 + X -> XSO4 + H2
"X" is any sutable metal. The volume of gas would be an indication of the number of moles, since at room temperature and pressure, 1 mole of any gas occupies 22.4 liters.
Proving that sulphuric acid reacting with MgCO3 is dibasic in a gas experiment how do I do this?
When a carbonate reacts with an acid, CO2 is produced. The ratio of acid used to CO2 produced is 1:1 for a diprotic acid, or 2:1 for a monoprotic acid, as the following equations demonstrate.
MgCO3 + H2SO4 -> MgSO4 + H2O + CO2 (diprotic acid)
MgCO3 + 2HCl -> MgCl2 + H2O + CO2 (monoprotic acid)
Collect the CO2 produced and compare with the quantity of acid needed. To produce a similar amount of CO2, you'll find that you need half the amount of H2SO4 compared to HCl.
