Try the new Google Books Check out the new look and enjoy easier access to your favorite features Page 2Solubility curve: The solubility of a given solute in water, as done above, can be determined at many different temperatures, say 0oC, 10oC, 20oC,…..100oC. The result so obtained can be plotted on a graph by taking temperatures along the X-axis, and solubility along the Y-axis. By joining the points so obtained, you get a curve, called the solubility curve. The solubility curves of six solutes in water are given in the graph. Observe the graph carefully and notice the following, regarding the solubility of these six solutes in water.
DEDUCTIONS of solubility curve:
For e.g., As per the given graph, the solubility of potassium nitrate at 60oC= 110 g and at 50oC = 91 g. Hence, by cooling a saturated solution of potassium nitrate containing 100 g of water from 60oC to 50oC, 19 g of the solute (110 g – 91 g) is deposited in the form of crystals. Applications of Solubility Curves
Calculations on Solubility Mathematically, solubility can be expressed as: Solubility (mol dm^3) = No. of moles/Volume (dm3)…………(i) = (No. of moles/Volume (cm3)) x 1000………(ii) Equations (i) and (ii) are used when the volume is given in dm^3 and cm^3 respectively. Also, Solubility (g dm^-3) = Reacting mass/Volume (dm3)…………(iii) = (Reacting mass/Volume (cm3)) x 1000………(iv) but, No. of moles = Reacting mass/Molar mass…………(v) Substituting equation (v) into (i) and (ii), then; Solubility (mol dm-3) = (Reacting mass/Molar mass) x 1000/Volume (cm3)……(vii) Equations (vi) and (vii) are used when the molar mass with the reacting mass is given. . Examples Question 1 Determine the solubility of salt Z in (i) g dm-3 (ii) mol dm-3 AnswerMass of saturated solution= Mass of dish + saturated solution of salt Z – Mass of empty dish= 35.70g – 14.32g = 21.38g Mass of salt Z = Mass of dish + salt Z – Mass of empty dish= 18.60g – 14.32g = 4.28g Molar Mass of salt Z = 100 Amount in moles of salt Z = Mass of salt Z/Molar mass of salt Z= 4.28/100 = 0.0428 mole Density of solution Z = 1.00g cm-3 Volume of solution = Mass of solution/Density of solution= 21.38g /1g cm-3 = 21.38 cm3 Therefore, solubility of salt Z in:i) g dm-3 = (Mass/Volume (cm3)) x 1000= (4.28/21.38) x 1000 = 200.19g dm-3
Question 2 AnswerMass of Pb(NO3)2 = 16.55gMolar mass of Pb(NO3)2 = (1xPb) + 2[(1xN) + (3xO)]= (1 x 207) + 2[(1 x 14) + (3 x 16)]= 207 + 2(14 + 48)= 207 + 124= 331g mol-1 Volume of water = 100cm3 (density of water = 1g cm-3) Solubility (mol dm-3) = (Reacting mass/Molar mass) x 1000/Volume (cm3)= (16.55/331) x (1000/100) = 0.05mol dm-3 Question 3 AnswerMolar mass of KClO3 = (1xK) + (1xCl) + (3xO)= (1 x 39) + (1 x 35.5) + (3 x 16)= 39 + 35.5 + 48 = 122.5g mol-1 Solubility of salt at 35°C = 1.8mol dm-3Solubility of salt at 20°C = 0.3mol dm-3Amount in moles of KClO3 crystallized out on cooling from 35°C – 25°C = 1.8 – 0.3 = 1.5 moles Therefore, mass of salt crystallized out on cooling = Amount in moles x Molar mass= 1.5 X 122.5 = 183.75g Question 4 AnswerSolubility of salt at 35°C = 0.35mol dm-3Molar mass of salt = 101gMass of salt in solution = 3.50g Volume of water = 250cm3 Solubility of salt in saturated solution = (Reacting mass/Molar mass) x 1000/Volume (cm^3)= (3.5/101) x (1000/250) = 0.138mol dm^-3 Since the calculated solubility of the salt solution in the beaker at 35°C, is less than the standard solubility of the salt at the same temperature, it means that the salt has not reached its solubility limit at that given temperature. Hence, the resulting solution will be unsaturated. Question 5 AnswerMolar mass of Na3AsO4.12H2O = (3xNa) + (1xAs) + (4xO) + 12[(2xH) + (1xO)]= (3 x 23) + (1 x 75) + (4 x 16) + 12[(2 x 1) + (1 x 16)]= 69 + 75 + 64 + 12(2 + 16)= 208 + 216 = 424g mol-1 From the above, molar mass of Na3AsO4 = 208g mol-1 Mass of Na3AsO4.12H2O in 100g of water = 38.9g Therefore, solubility of Na3AsO4.12H2O (g dm-3)= (Mass/Volume (cm3 )) x 1000= (38.9/100) x 1000 = 389g dm-3 but, 424g of Na3AsO4.12H2O contains 208g of Na3AsO4Therefore,389g of Na3AsO4.12H2O will contain x of Na3AsO4x = (389 x 208)/424 = 190.83g of Na3AsO4 Percentage of Na3AsO4 in the saturated solution = 49.1% Question 680g of a salt, XCl2, was placed in 40cm3 of water to give a saturated solution at 25°C. If the solubility of the salt is 8 mol dm-3 at that temperature, what is the mass of the salt that will be left undissolved at the given temperature? [X = 24, Cl = 35.5] AnswerMolar mass of XCl2 = (1xX) + (2xCl)= (1 x 24) + (2 x 35.5)= 24 + 71= 95g mol-1At 25°C,Solubility of XCl2 = 8.0mol dm-3 i.e, 1 dm3 of solution contains 8 x 95 = 760g If, 1000cm3 of water = 760g of XCl2Then, 40cm3 of water = x g of XCl2x = 760 x 40/1000 = 30.4g of XCl2 From the above, based on the solubility of the salt, 40cm3 will only dissolve 30.4g Therefore, the mass of the salt that will be left undissolved is 80 – 30.4 = 49.6g Question 7 AnswerMasses at 80°C, 1000g of water + 1500g of KClO3 = 2500g of saturated solution Masses at 45°C, Mass of salt deposited on cooling from 80°C to 45°C If 2500g of saturated solution deposit 900g of solute on coolingThen, 200g of the saturated solution will deposit x g of solutex = 900 x 200/2500 = 72g of KClO3. Do These Question 1 Question 2 Question 3 Question 4 Question 5a) What do you understand by terms saturated solution and solubility? b) When 50cm3 of a saturated solution of sugar of molar mass 345g at 50°C was evaporated to dryness, 34.5g of dry solid was obtained. Calculate the solubility of sugar ay 50°C in a) g per dm3 b) mol per dm3 Question 6 REVISION EXERCISES (POST ANSWERS USING QUESTION BOX BELOW FOR EVALUATION AND DISCUSSION. ADD QUESTION TITLE)
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