Example : 100 cm3 of a magnesium hydroxide solution required 4.5 cm3 of sulphuric acid (of concentration 0.1 mol dm-3) for complete neutralisation. [atomic masses: Mg = 24.3, O = 16, H = 1)
(a) give the equation for the neutralisation reaction. (b) calculate the moles of sulphuric acid neutralised. (c) calculate the moles of magnesium hydroxide neutralised. (d) calculate the concentration of the magnesium hydroxide in mol dm-3 (molarity). (e) calculate the concentration of the magnesium hydroxide in g cm-3
a) Mg(OH)2(aq) + H2SO4(aq) ==> MgSO4(aq) + 2H2O(l)
(b) moles of sulphuric acid neutralised = 0.1 x 4.5/1000 = 0.00045 mol
(c) moles of magnesium hydroxide neutralised also = 0.00045 (1:1 in equation) in 100 cm3
(d) concentration of the magnesium hydroxide in mol dm-3 = 0.00045 x 1000 ÷ 100 = 0.0045 (scaling up to 1000cm3=1dm3, to get molarity)
(e) molar mass of Mg(OH)2 = 58.3 so concentration of the magnesium hydroxide = 0.0045 x 58.3 = 0.26 g dm-3 (= g per 1000 cm3),so concentration = 0.26 ÷ 1000 = 0.00026 g cm-3
Example : A bulk solution of hydrochloric acid was standardised using pure anhydrous sodium carbonate (Na2CO3, a primary standard). 13.25 g of sodium carbonate was dissolved in about 150 cm3 of deionised water in a beaker. The solution was then transferred, with appropriate washings, into a graduated flask, and the volume of water made up to 250 cm3, and thoroughly shaken (with stopper on!) to ensure complete mixing.
25 cm3 of the sodium carbonate solution was pipetted into a conical flask and screened methyl orange indicator added. The aliquot required 24.65 cm3 of a hydrochloric acid solution, of unknown molarity, to completely neutralise it. [atomic masses: Na = 23, C = 12, O = 16]
(a) Calculate the molarity of the prepared sodium carbonate solution.(b) Write out the equation between sodium carbonate and hydrochloric acid, including state symbols. (c) How many moles of sodium carbonate were titrated? (d) How many moles of hydrochloric acid were used in the titration? (e) What is the molarity of the hydrochloric acid?
(a) moles = mass/f. mass, f. mass Na2CO3 = 106, mol Na2CO3 = 13.25/106 = 0.125
molarity = mol/vol. in dm3, 250 cm3 = 0.25 dm3,molarity Na2CO3 = 0.125/0.25 = 0.50 mol dm-3
(b) Na2CO3(aq) + 2HCl(aq) ==> 2NaCl(aq) + H2O(l) + CO2(g)
(c) mol = molarity x volume
(d) mol Na2CO3 titrated = 0.5 x 25/1000 = 0.0125 mol Na2CO3 (in the 25 cm3 aliquot pipetted)
from equation, mole ratio Na2CO3:HCl is 1:2,
so mol HCl = 2 x mol Na2CO3 = 2 x 0.0125 = 0.025 mol HCl (in the 24.65 cm3 titre)
(e) molarity = mol/vol. in dm3, dm3 = cm3/1000, 24.65/1000 = 0.02465 dm3 therefore molarity HCl = 0.025/0.02465 = 1.014 mol dm-3 (1.01 3sf)
Example : (a) Describe a procedure that can used to determine the molecular mass of an organic acid by titration with standardised sodium hydroxide solution. Indicate any points of the procedure that help obtain an accurate result and explain your choice of indicator.
0.279g of an organic monobasic aromatic carboxylic acid, containing only the elements C, H and O, was dissolved in aqueous ethanol. A few drops of phenolphthalein indicator were added and the mixture titrated with 0.100 mol dm-3 sodium hydroxide solution. It took 20.50 cm3 of the alkali to obtain the first permanent pink. [at. masses: C = 12, H = 1 and O = 16]
(b) How many moles of sodium hydroxide were used in the titration? (c) How many moles of the organic acid were titrated? and explain your reasoning. (d) Calculate the molecular mass of the acid. (e) Suggest possible structures of the acid with your reasoning
(a) An appropriate quantity of the acid is weighed out, preferably on a 4 sf electronic balance. It can be weighed into a conical flask by difference i.e. weight acid added to flask = (weight of boat + acid) - (weight of boat). The acid is dissolved in water, or aqueous-ethanol if not very soluble in water. The solution is titrated with standard sodium hydroxide solution using phenolphthalein indicator until the first permanent pink. The burette should be rinsed with the sodium hydroxide solution first. The flask should be rinsed around the inside to ensure all the acid and alkali react, and drop-wise addition close to the end-point to get it to the nearest drop.
The pKind for phenolphthalein is 9.3, and its effective pH range is 8.3 to 10.0. The pH of a solution of the sodium salt of the acid (from strong base + weak acid) is in this region and so the equivalence point can be detected with this indicator. (page on acid-base equilibria and theory of indicators is in production)
(b) moles = molarity x volume in dm3 (dm3 = cm3/1000) mol NaOH = 0.10 x 20.5/1000 = 0.00205 mol
(c) mol NaOH = mol RCOOH = 0.00205 because 1:1 mole ratio for a monobasic acid: RCOOH + Na+OH- ==> RCOO-Na+ + H2O
(d) moles = mass (g)/Mr, so Mr = mass/mol = 0.279/0.00205 = 136.1
(e) The simplest aromatic carboxylic acid is benzoic acid C6H5COOH, Mr = 122， 136-122 = 14, which suggests an 'extra' CH2 (i.e. -CH3 attached to the benzene ring instead of a H), so, since the COOH is attached to the ring, there are three possible positional/chain isomers of CH3C6H4COOH (Mr = 136) 2-, 3- or 4-methylbenzoic acid.
moles Z = mass of Z gas (g) / atomic or formula mass of gas Z (g/mol)
gas volume of Z = moles of Z x molar volume
Example 9.6: A small teaspoon of sodium hydrogencarbonate (baking soda) weighs 4.2g. Calculate the moles, mass and volume of carbon dioxide formed when it is thermally decomposed in the oven. Assume room temperature for the purpose of the calculation.
2NaHCO3(s) ==> Na2CO3(s) + H2O(g) + CO2(g)
Formula mass of NaHCO3 is 23+1+12+(3x16) = 84 = 84g/mole
Formula mass of CO2 = 12+(2x16) = 44 = 44g/mole (not needed by this method)
or a molar gas volume of 24000 cm3 at RTP (definitely needed for this method)
In the equation 2 moles of NaHCO3 give 1 mole of CO2 (2:1 mole ratio in equation)
Moles NaHCO3 = 4.2/84 = 0.05 moles ==> 0.05/2 = 0.025 mol CO2 on decomposition.
Mass = moles x formula mass, so mass CO2 = 0.025 x 44 = 1.1g CO2
Volume = moles x molar volume = 0.025 x 24000 = 600 cm3 of CO2
Avogadro's Law states that 'equal volumes of gases at the same temperature and pressure contain the same number of molecules' or moles of gas
Example : C3H8(g) + 5O2(g) ==> 3CO2(g) + 4H2O(l)
(a) What volume of oxygen is required to burn 25cm3 of propane, C3H8.
Theoretical reactant volume ratio is C3H8 : O2 is 1 : 5 for burning the fuel propane.
so actual ratio is 25 : 5x25, so 125cm3 oxygen is needed.
(b) What volume of carbon dioxide is formed if 5dm3 of propane is burned?
Theoretical reactant-product volume ratio is C3H8 : CO2 is 1 : 3
so actual ratio is 5 : 3x5, so 15dm3 carbon dioxide is formed.
(c) What volume of air (1/5th oxygen) is required to burn propane at the rate of 2dm3 per minute in a gas fire?
Theoretical reactant volume ratio is C3H8 : O2 is 1 : 5
so actual ratio is 2 : 5x2, so 10dm3 oxygen per minute is needed,
therefore, since air is only 1/5th O2, 5 x 10 = 50dm3 of air per minute is required
You can use the ideas of relative atomic, molecular or formula mass AND the law of conservation of mass to do quantitative calculations in chemistry.
On analysis, a sample of hard water was found to contain 0.056 mg of calcium hydrogen carbonate per cm3 (0.056 mg/ml). If the water is boiled, calcium hydrogencarbonate Ca(HCO3)2, decomposes to give a precipitate of calcium carbonate CaCO3, water and carbon dioxide.
(a) Give the symbol equation of the decomposition complete with state symbols.
Ca(HCO3)2(aq) ==> CaCO3(s) + H2O(l) + CO2(g)
(b) Calculate the mass of calcium carbonate in grammes deposited if 2 litres (2 dm3, 2000 cm3 or ml) is boiled in a kettle. [ atomic masses: Ca = 40, H = 1, C = 12, O = 16 ]
the relevant ratio is based on: Ca(HCO3)2 ==> CaCO3
The formula masses are 162 (40x1 + 1x2 + 12x2 + 16x6) and 100 (40 + 12 + 16x3) respectively
there the reacting mass ratio is 162 units of Ca(HCO3)2 ==> 100 units of CaCO3
the mass of Ca(HCO3)2 in 2000 cm3 (ml) = 2000 x 0.056 = 112 mg
therefore solving the ratio 162 : 100 and 112 : z mg CaCO3
where z = unknown mass of calcium carbonate
z = 112 x 100/162 = 69.1 mg CaCO3
since 1g = 1000 mg, z = 69.1/1000 = 0.0691 g CaCO3, calcium carbonate
(c) Comment on the result, its consequences and why is it often referred to as 'limescale'?
This precipitate of calcium carbonate will cause a white/grey deposit to be formed on the side of the kettle, especially on the heating element. Although 0.0691 g doesn't seem much, it will build up appreciably after many cups of tea! The precipitate is calcium carbonate, which occurs naturally as the rock limestone, which dissolved in rain water containing carbon dioxide, to give the calcium hydrogen carbonate in the first place. Since the deposit of 'limestone' builds up in layers it is called 'limescale'.
Posts
Blog relocation
17 years ago
