Concentration and units

Mixtures do not have a defined composition so we need to define concentration to know the proportion between solute and solvent.

Concentration is every magnitude that expresses the ratio between theamount of solute and solvent in a solution. This can be solid in liquid, liquidin liquid, solidin solid (alloys), gas in liquid, solidin gas, liquidin solid, etc…

There are many different concentration units, each with their own for and against:

-g/L: grams of solute per litre of solution

-% (w/w): weight percentageof solute with respect to the total weight of the solution; it usually comes with the density of the solution

-Mor mol/L: molarity, number of moles of solute per litreof solution; this is by excellence the Chemistry unit

-m: molality, number of moles of solute per kilogram of solvent

-x: molar fraction, number of moles of solute divided by the total number of moles of the solution

Richness: relative abundance of the component of interest in the sample.

Purity: percentage of the component of interest in the mixture.

Colligative properties of solutions

Colligative properties are those which depend on the concentration of the solution. These properties do not depend on the nature or type of solute but only the amount of particles of solute present in the solution. This is the reason why we will only consider non-dissociable solutes, in which the concentration directly indicates the number of particles.

The colligative properties are four: vapour pressure lowering, freezing point depression, boiling point elevation and osmotic pressure.

1.-Vapour pressure lowering: Raoult’s law

In 1887, Raoultfound experimentally that partial vapour pressure of solvent, when in presence of a non-volatile solute, is directly proportional to the molar fraction of the solvent:

p = p0·xD

beingpthe vapour pressure of the solvent in the solution, p0, that of pure solvent and xDthe molar fraction of the solvent. The variation in vapour pressure will therefore be:

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Figure 2.The presence of a solute lowers down the vapour pressure of (a) a solutionin comparison with (b) the pure compound, this is, less molecules evaporate to the gas phase.

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The molar fraction of a pure solvent is, logically, 1, and therefore the vapour pressure is that of the pure compound. If it is not pure, as it happens with solutions,the vapour pressure of the solvent can only be lower.

2. Freezing point depression

This is the lowering in the freezing point with respect to that of the pure solvent, given by:

Tc = Kc·m

whereTbis the depression in the freezing point, Kc, the cryoscopic constant, andm,the molality. For water Kc = 1.86°C kg/mol.

3. Boiling point elevation

This is the increase in the temperature of the boiling point with respect to that of the pure solvent, given by:

Tb = Kb·m

whereTbis the elevation in the boiling point, Kb, the ebullioscopic constantof the solvent, andm, the molality. For water, Kb = 0.52°C kg/mol.

The previous expressions are only valid for non-dissociable, non-ionisable solutes. In the case of ionisable solutes the expression should be modified and the corresponding ionisation degree factor introduced.

HOMO FABER: Antifreezers

Man has learnt to explode on his own benefit the freezing-point depression in order to avoid dangerous winter situations. For example, cars have a water closed circuit to cool down engines while working. If outside temperatures go below 0 °C, this water could freeze, increasing its volume and pressing on every conduction, and could make the whole circuit explode. To prevent this circumstance, water is replaced by a very concentrated solution so that its freezing point is much lower than that of the water.

Another use of cryoscopy is the possibility of making ice disappear from roads by spreading salt. Salt dissolves in the very thin liquid film ice holds, producing salty water, which at 0 °Cis liquid. As the resulting solution is very concentrated, ice tends to melt in order to dilute it until there is no more ice or the solution is so diluted that its freezing points does not differ from that of the pure water, and it freezes. Anyway, when salty water freezes it forms a snow-like solid, not ice, so the slipping problem is overtaken.

4. Osmotic pressure

The level of liquid in two communicating vessels should reach the same height. However, if we locate a semi-permeable membrane in between, only the solvent will be able to go across the separation to the more concentrated solution, trying to balance both concentrations, and reaching so the equilibrium situation in which both solutions have the same vapour pressure. This process is called osmosis.

Figure 3. Types of membranes according to their permeability.

If we put the same level of solution and pure solvent in both branches of a U-tube, separated by a semi-permeable membrane in the middle, the osmotic process will make the branch with the solution go higher due to the transference of solvent from the other one, which will subsequently descend. This difference in height generates a hydrostatic pressure over the solvent through the membrane. This pressure is what we call osmotic pressure.

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Van’t Hoffhad the idea of the particles of solute in the solution behaving similarly to an ideal gas, so the exerted pressure would be:

·V = n·R·T

where donde is the osmotic pressure in atm, Vis the volume of solution, n, number of moles of solute, R, the constant of gases and Tthe temperaturein Kelvin. Working out , and given that n/V is M, the molar concentration is:

 = (n/V)·R·T = M·R·T

As we have said that this is a hydrostatic pressure, then,

 = ·g·h

·g·h = (n/V)·R·T

Osmotic pressureon a solution is the pressure exerted by the solvent through the membrane, and balances the hydrostatic pressure corresponding to the difference in height between both columns.

HEALTH AND SOCIAL EDUCATION: Physiological media

You may have wondered about what is contained in the hanging solutions we can see in hospitals. This is not medicine (although these can be dissolved in these bags to make easier their administration), butsalt water.

But why salt water? Well, blood contains lots of different cells, mostly red blood cells; a red blood cell is made of a cytoplasm (quite a concentrated solution) separatedfrom plasma by a membrane (which lets substances selectively go through, this is, semipermeable). If we put red blood cells in distilled water (hypotonic medium), water begins to get into the cell to try and dilute the inside solution, finally bursting the cell out. On the contrary, if we put the red blood cells in a very concentrated solution (hypertonic medium), then the water goes out from the cells to try and dilute the surrounding solution, “drying out” themselves. In both cases the final result is the death of the cell, therefore showing how important is that the infused solutions have a concentration as similar as possible to that of the physiological medium, an isotonicsolution.

Stoichiometry.

Limiting and excess reagents

We call limiting reagent to that one that is firstly consumed, stopping the reaction, and therefore limiting the amount of product that can be obtained. In opposition, we callreagent in excess to the rest of reagents in the reaction. In order to work out which one is the limiting reagent there is avery simple technique: we just have to express the amount of every reagent in moles and divide these quantities by the corresponding stoichiometric coefficient; the smallest one indicates which is our limiting reagent. In order to determine how much is this excess, it has to be worked out how much would react with he limiting reagent, and subtract it from the total. Once defined the limiting reagent, all stoichiometric calculations will be performed taking this as a reference.

Reaction yield

In spite of the common use, chemical reactions do not have to occur in their whole extension, but can stop before all reagents have been consumed. We call reaction yield to the percentage of reacted quantitywith respect to the total one (it could also be defined as the actually produced with respect to the theoretically possible amount of product, as a percentage).

Questions and problems

  1. Determine the molarity of a sulphuric acid solution, obtained by addition of 196 gof the acid on water until 2 Lof solution are obtained.
  2. How many grams of potassium hydroxide will we have to weigh to obtain 250 cm3of 2 Msolution?
  3. Determine the molality of a solution of nitric acid, if63 gof acid are added to 500 cm3of water.
  4. Calculatethe molar fraction of every componentin the following solutions: a) 2 gof acetic in 15.6 gof benzene; b) 11.7 gof common salt in 144 gof water.
  5. The density of a hydrochloric acid solution is 1.101 g/mLand contains 19.91 % of HCl. Calculate its molarity, molality and molar fraction of HCl in the solution.
  6. A solution is prepared by dissolving 9.75 gof ferric chloride in 500 mLof water. Calculate: a) molarity, b) molality, c) molar fractionof each component, d) number of moles of ferric chloride in 5 mLof solution, e) number of iron atoms in this volume, and f) total number of ions.
  7. Calculate how much ethanol would be neededper litre of water in order not to freezeabove-4.8°C.
  8. Commercial sulphuric acidhas a density of 1.8 g/mLand richness in weight of 98 %. Calculate its molarityand molality. What will be the corresponding vapour pressure of water for the solution at20 °C? Pv(water, 20 °C)= 18 Torr.
  9. 20.7 gof ethanol are dissolved in100 gof water, resulting in a volume of 112 mL. Calculatethe freezing and boiling points of the solution prepared. How much will the osmotic pressure be at 25 °C?
  10. An aqueous solution of 9.2 g/Lof a certain substance exerts an osmotic pressure of 0.474 atm at0°C. Calculatethe molecular mass of the solute.
  11. How many grams of acetic acid should we dissolve in 250 mLof water in order to change its boiling point up to 101.3°C? What would the corresponding freezing point?
  12. When dissolving 5 gof a certain solute in 50 gof water, the resulting solution boils at 100.5°C. What is the corresponding molecular mass of the solute?
  13. What is the vapour pressure of problem 11 at 20 °C? And what should it be at100 °C? At what temperature will this solution boil?
  14. Explain with your own words how antifreezers work.
  15. 4.9 gof zinc react with 25 mLof0.02 Mhydrochloric acid, giving up zinc chloride and molecular hydrogen. Calculate which is the limiting reagent, the reagent in excess, the excess and how many litres of hydrogen will be released at 684 Torr and35 °C. Data: Zn 65.
  16. 30 gof calcium carbonate,richness 70 %, react with 50 mLof sulphuric acid 0.1 M, yielding calcium sulphate, water and carbon dioxide. Calculate the limiting reagent, reagent in excess, excess, number of molecules of water produced and volume of carbon dioxide released at20°Cand 900 Torr.
  17. Mercury(II) oxide thermally decomposes into mercury and oxygen. Given a reaction yield of 80 %, how much oxide was used if 2 Lof oxygen were released at 273°Cand646 Torr? Datos: Hg 200.5.
  18. 216 gof aluminium are made to react with hydrochloric acid. Calculate a) the number of grams of salt obtained, and b) the number of spent moles of acid.Data: Al 27
  19. In a given reaction 250 gof calcium carbonate were spent when in contact with perchloric acid. Calculate a) the number of grams obtained of the corresponding salt, and b) the volume of carbon dioxide released at 912 Torr and250 °C. Data: Ca 40
  20. How many molecules of water will be obtained in the reaction of 250 mL of 3 M sulphuric acid solution with potassium hydroxide? How many grams of potassium hydroxide are needed for this reaction?
  21. 42 gof calcium oxide 80 % rich, are made to react with 2.5 Lof 0.25 Mperchloric acid, giving up water and calcium perchlorate. Calculate the limiting reagent, reagent in excess, excess and number of grams of calcium perchlorate formed.Data: P 31, Ca 40.
  22. 12 gof calcium chloride, richness 74 %, react with 300 mLnitric acid0.5 M, producing calcium nitrate and hydrogen chloride, which is released as a gas. Determinethe limiting reagent, reagent in excess, excess and number of grams of nitrate formed, and the pressure exerted by the hydrogen chloride produced if it were introduced in a 1Lflask at70°C.
  23. 6 gof Mg with richness of 80 % react with 25 mLof hydrochloric acid 37.1 %rich and a density of 1.18 g/mL. Determine the limiting reagent, reagent in excess, excess, and volume of hydrogen released at 16 °Cand 798 Torr. Data: Mg 24
  24. 250 mL of 0.25 Mhydrofluoricacid react with 4.5 gof calcium carbonate, richness 85 %, to produce calcium fluoride, carbon dioxide and water. Determine the limiting reagent, reagent in excess, excess in grams, volume of carbon dioxide produced at 30 °Cand 500 mmHg, and number of water molecules generated.
  25. 400 mLof0.3 M ammonia solution react with 4.92Lof oxygen, measured at 27°Cand 0.8 atm, producing nitrogen(II) oxide and water. Calculate which is the limiting reagent, reagent in excess, excess in grams, volume of nitrogen(II) oxide measured in the previously mentioned conditions, and number of grams of water generated.

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