Concentration of a solution



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Revision of calculations Stoichiometric calculations Osmotic pressure and osmolarity MUDr. Jan Pláteník, PhD Concentration of a solution mass concentration: grams of substance per litre of solution molar concentration: moles of substance per litre of solution in %: % (w/v): weight per volume, grams of substance per 100 ml of solution % (v/v) volume per volume, ml of substance per 100 ml of solution 1

Molar Volume one mole of any gaseous substance occupies the same volume at the same temperature and pressure..22.414 litres at 101.325 kpa, 0 C (273.15 K) (Avogadro s Law) P: pressure in kpa V: volume in dm 3 (l) n: number of moles P. V = n. R. T R: universal gas constant (8.31441 N.m.mol -1.K -1 ) T: temperature in K 2

Example: what is volume of one mole of gas at 101.325 kpa and 25 C? P.V = n.r.t n.r.t 1 x 8.31441 x (273.15+25) V = = = P 101.325 = 24.465 dm 3 Conversions between mass and molarity: Summary Always distinguish between amount of substance in moles (grams) and concentration of substance in mol/l (g/l) For conversion from mass to molarity divide the mass (g or g/l) with molar mass (relative AW/MW/FW) For conversion from molarity to mass multiply the molarity (mol or mol/l) with molar mass (relative AW/MW/FW) 3

Conversion from mass to molar Example: Calculate molar concentration of Na 2 HPO 4 solution c = 21 g /l. (AW of Na: 23, P: 31, O: 16, H: 1) FW of Na 2 HPO 4 : 46+1+31+4x16 = 142 Molar concentration = Mass conc. (g/l) / FW = 21 / 142 = 0.15 mol/l Conversion from molar to mass Example: Calculate how many g of KClO 4 is needed for preparation of 250 ml of 0.1 M solution. (AW of K: 39, Cl: 35.4, O: 16) FW of KClO 4 : 39 + 35.4 + 4x16 = 138.4 Mass conc. = molar conc. x FW we need 138.4 x 0.1 x 0.25 = 3.46 g KClO 4 4

Conversion from % to molarity Example: The physiological saline is NaCl 0.9 % (w/v) What is molar concentration of NaCl in this solution? (AW of Na: 23, Cl: 35.5) FW of NaCl : 23+35.45 = 58.5 0.9 % (w/v) is 0.9 g/100 ml = Mass conc. 9 g/l Molar concentration = Mass conc. (g/l) / FW = 9/58.5 = 0.154 mol/l Calculations with molar volume Example: What is weight (in grams) of 1 liter of oxygen at atmospheric pressure and ambient temperature? (AW of O: 16) Molar volume at 101.325 kpa and 25 C: 24.5 l/mol 1 liter of oxygen is 1/24.5 = 0.040816 mol Conversion to mass: 0.040816 x 32 = 1.306 g 5

Diluting solutions Example: You need to prepare 1 liter of 0.1 M HCl. How many ml of concentrated HCl (12 M) do you need to take? c 1. v 1 = c 2. v 2 12 x v 1 = 0.1 x 1000 v 1 = 100/12 = 8.33 ml What is molarity of pure water? Molar concentration: moles of substance per liter of solution 1 liter of water weighs 997 g at 25 C FW of H 2 O: 2+16=18 997 g H 2 O is 997/18 = 55.4 moles Molarity of pure water is 55.4 mol/l 6

Stoichiometric calculations Titration calculations Example: An unknown sample of sulfuric acid H 2 SO 4 was titrated with the known KOH solution. It was found that 12 ml of the KOH c=0.1 mol/l was needed for just complete neutralisation of 10 ml H 2 SO 4 unknown sample. What is concentration of sulfuric acid in the sample? Equation: H 2 SO 4 + 2 KOH K 2 SO 4 + 2 H 2 O Calculation: H 2 SO 4 KOH c 1. v 1 = c 2. v 2 c 1 = c 2. v 2 / v 1 c 1 = 0.1. 12 / 10 = 0.12 Including stoichiometry : c(h 2 SO 4 ) = 0.12/2 = 0.06 mol/l 7

Stoichiometric calculations Example: In the reaction between barium nitrate and sodium sulfate, how many grams of barium sulfate can be prepared from 10 ml of 10 % (w/v) barium nitrate? Take into account that about 5% of the product is lost. (AW of barium: 137.3, sulfur: 32.1, nitrogen: 14.0, oxygen: 16.0) equation: Ba(NO 3 ) 2 + Na 2 SO 4 BaSO 4 + 2NaNO 3 FW Ba(NO 3 ) 2 : 261.3 FW BaSO 4 : 233.4 10 ml of 10% (w/v) Ba(NO 3 ) 2 : 1 g... 1/261.3 = 0.003827 moles amount of BaSO 4 formed: 0.003827 moles...... 0.003827 x 233.4 = 0.8932 g (theoretical yield, 100%) Actual yield: 0.8932 x 0.95 = 0.849 g Stoichiometric calculations Example II: Propane gas burns on air to water and carbon dioxide: CH 3 CH 2 CH 3 (g) + O 2 (g) H 2 O(g) + CO 2 (g) Balance the equation and then calculate how many grams of water and carbon dioxide will result from burning of 10 liters of propane gas (molar volume at 25 C and atmospheric pressure: 24.465 l/mol ) Balanced equation: CH 3 CH 2 CH 3 (g) + 5O 2 (g) 4H 2 O(g) + 3CO 2 (g) 10 l? (MW: 18)? (MW: 44) 10 l of propane: 0.4087 moles... 1.635 mol H 2 O = 29.4 g... 1.226 mol CO 2 = 53.9 g 8

Stoichiometric calculations Example III: Solid lithium hydroxide is used in space shuttles to remove exhaled carbon dioxide from the living environment: 2LiOH(s) + CO 2 (g) Li 2 CO 3 (g) + H 2 O(l) Imagine that you are planning a space mission of two astronauts for 72 hours. One astronaut will produce 250 ml of CO 2 per minute at rest. How many kg of solid lithium hydroxide are needed for the mission? (AW Li: 6.941, molar volume 24.465 L/mol) CO 2 production: 0.25 L 60 72 2 = 2160 L (88.3 mol) 2 88.3 mol LiOH needed: 176.6 mol FW LiOH: 6.941 + 17 = 23.941 176.6 mol LiOH is 4228 g = ~ 4.23 kg Osmolarity Osmotic pressure 9

Osmolarity Sum of all osmotically active particles (OAP) in a solution One of the colligative properties: only number of particles counts, not their kind! Units: moles of osmotically active particles (osm) per liter Example: NaCl 0.15 mol/l, in solutions exists as ions Na +, Cl 2 OAP.. Osmolarity 0.3 mol OAP/l (0.3 osm/l) Osmotic pressure Π = n. C. R. T Osmolarity Figure from: http://www.bbc.co.uk/dna/h2g2/a686766 Π: osmotic pressure in kpa n: number of OAP per mole C: molar concentration in mol/l R: universal gas constant (8.31441 N.m.mol -1.K -1 ) T: temperature in K 10

Osmolarity is important in medicine Isotonic: osmolarity = 0.3 osm/l (NaCl 0.15 mol/l) Hypotonic: osmolarity < 0.3 osm/l Hypertonic: osmolarity > 0.3 osm/l Figure from: Wikipedia Calculations of osmolarity/osmotic pressure Example I: Calculate osmolarity of Na 2 HPO 4 solution of c = 21 g /l. (AW of Na: 23, P: 31, O: 16, H: 1) FW of Na 2 HPO 4 : 46+1+31+4x16 = 142 Molar concentration = Mass conc. (g/l) / FW = 21 / 142 = 0.15 mol/l 3 OAP: 2x Na +, 1x HPO 4 2- Osmolarity: 0.15 x 3 = 0.45 mol OAP/l 11

Calculations of osmolarity/osmotic pressure Example II: Calculate osmotic pressure of Na 2 HPO 4 solution of c = 21 g /l at 37 C. (AW of Na: 23, P: 31, O: 16, H: 1; R = 8.31441 N.m.mol -1.K -1 ) FW of Na 2 HPO 4 : 46+1+31+4x16 = 142 Molar concentration 3 OAP: 2x Na +, 1x HPO 4 2- Osmolarity: 0.15 x 3 = 0.45 mol OAP/l Osmotic pressure: = Mass conc. (g/l) / FW = 21 / 142 = 0.15 mol/l Π = n. C. R. T Π = 3 x 0.15 x 8.31441 x (273.15 + 37) = 1160 kpa (hypertonic) Calculations of osmolarity/osmotic pressure Example III: 0.2 M KCl solution is combined with an equal volume of 0.5 M glucose solution. What is the resulting osmolarity? After mixing: KCl: c= 0.1 mol/l, 2 OAP: contribution to osmolarity 0.2 mol/l Glucose: c=0.25 mol/l, 1 OAP: contribution to osmolarity 0.25 mol/l Total osmolarity: 0.2 + 0.25 = 0.45 mol OAP/l 12

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