H co 3 2 has the higher charge density because it has

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h) CO 3 2– has the higher charge density because it has both a smaller ion volume and greater charge. 13.33 a) I has a smaller charge density (larger ion volume) than Br . b) Ca 2+ has a lower ratio than Sc 3+ , due to its smaller ion charge. c) Br has a lower ratio than K + , due to its larger ion volume. d) Cl has a lower ratio than S 2– , due to its smaller ion charge. e) Sc 3+ has a lower ratio than Al 3+ , due to its larger ion volume. f) ClO 4 has a lower ratio due to its smaller ion charge. g) Fe 2+ has a lower ratio due to its smaller ion charge. h) K + has a lower ratio due to its smaller ion charge. 13.34 Plan: The ion with the greater charge density will have the larger H hydration . Solution: a) Na + would have a larger H hydration than Cs + since its charge density is greater than that of Cs + . b) Sr 2+ would have a larger H hydration than Rb + . c) Na + would have a larger H hydration than Cl . d) O 2– would have a larger H hydration than F . e) OH would have a larger H hydration than SH . f) Mg 2+ would have a larger H hydration than Ba 2+ . g) Mg 2+ would have a larger H hydration than Na + . h) CO 3 2– would have a larger H hydration than NO 3 . 13-5
13.35 a) I b) Ca 2+ c) Br d) Cl e) Sc 3+ f) ClO 4 g) Fe 2+ h) K + 13.36 Plan: Use the relationship H solution = H lattice + H hydration . Given H solution and H lattice , H hydration can be calculated. H hydration increases with increasing charge density, and charge density increases with increasing charge and decreasing size. Solution: a) The two ions in potassium bromate are K + and BrO 3 . H solution = H lattice + H hydration H hydration = H solution H lattice = 41.1 kJ/mol – 745 kJ/mol = –703.9 = –704 kJ/mol b) K + ion contributes more to the heat of hydration because it has a smaller size and, therefore, a greater charge density. 13.37 a) H hydration = H solution H lattice H hydration = 17.3 kJ/mol –763 kJ/mol H hydration = –745.7 = – 746 kJ/mol b) Na + ion contributes more to the heat of hydration due to its smaller size (larger charge density). 13.38 Plan: Entropy increases as the possible states for a system increase, which is related to the freedom of motion of its particles and the number of ways they can be arranged. Solution: a) Entropy increases as the gasoline is burned. Gaseous products at a higher temperature form. b) Entropy decreases as the gold is separated from the ore. Pure gold has only the arrangement of gold atoms next to gold atoms, while the ore mixture has a greater number of possible arrangements among the components of the mixture. c) Entropy increases as a solute dissolves in the solvent. 13.39 a) Entropy increases . b) Entropy decreases . c) Entropy increases . 13.40 H solution = H lattice + H hydration H solution = 822 kJ/mol + (– 799 kJ/mol) H solution = 23 kJ/mol 13.41 Add a pinch of the solid solute to each solution. A saturated solution contains the maximum amount of dissolved solute at a particular temperature.

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