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Science Chemistry Chemistry: An Atoms-Focused Approach (Second Edition)

The concentration of N i N O 3 2 in the given solution The concentration of N i 2 + ( a q ) in the given solution Solution: N i N O 3 2 = 4.00 × 10 - 3 M N i 2 + = 6.3 × 10 - 10 M Explanation: Concept: Initial concentration of N i N O 3 2 can be calculated using its given moles and the volume of the solution. After addition of the 0.500 M N H 3 solution into N i ( N O 3 ) 2 solution, there would be the formation of a [ N i ( N H 3 ) 6 ] 2 + complex. This complex would be in the equilibrium with N i 2 + , so by using the formation constant equation, we can find out the concentration of N i 2 + at equilibrium. Formula: m o l a r i t y = m o l e s o f s o l u t e v o l u m e o f s o l u t i o n i n L K f = [ N i ( N H 3 ) 6 ] 2 + N i 2 + [ N H 3 ] 6 Given: M o l e s o f N i N O 3 2 = 1 m m o l × 1 m o l 1000 m m o l = 0.001 m o l v o l u m e o f s o l u t i o n = 250 m L × 1 L 1000 m L = 0.250 L N H 3 = 0.500 M Calculations: a) Calculation for the concentration of N i N O 3 2 in the solution: m o l a r i t y = m o l e s o f s o l u t e v o l u m e o f s o l u t i o n i n L m o l a r i t y = 0.001 m o l 0.250 L = 0.004 M = 4.00 × 10 - 3 M b) Calculation for the concentration of N i 2 + ( a q ) in the solution: The complex formation reaction is N i 2 + a q + 6 N H 3 a q ⇌ N i N H 3 6 2 + ( a q ) Using reaction stoichiometry, we calculate the consumed mole of N H 3 . 1 m o l N i ( N O 3 ) 2 reacts with 6 m o l N H 3 , so 0.00100 m o l N i ( N O 3 ) 2 will react with 0.00100 m o l N i ( N O 3 ) 2 × 6 m o l N H 3 1 m o l N i ( N O 3 ) 2 = 6 × 10 - 3 m o l N H 3 We can see that N H 3 is an excess reagent, so the remaining moles of ammonia after the reaction is 0.500 M N H 3 - 6 × 10 - 3 M N H 3 = 0.476 M N H 3 Since K f is large, we can assume that essentially all the N i 2 + ions are converted to complex ions and N i N H 3 6 2 + = 0.00400 M . A negligible concentration of free N i 2 + ions remains at equilibrium which is assumed as x . We can calculate the [ N i 2 + ] by considering how much N i ( N H 3 ) 6 dissociates: N i N H 3 6 2 + ⇌ N i 2 + a q + 6 N H 3 ( a q ) The equilibrium constant, K d for this reaction is the reciprocal of K f . K d = 1 K f = N i 2 + [ N H 3 ] 6 [ N i ( N H 3 ) 6 ] 2 + = 1 5.5 × 10 8 = 1.82 × 10 - 9 We can construct a RICE table incorporating the concentrations and concentration changes of the products and reactants. Reaction N i N H 3 6 2 + ( a q ) ⇌ N i 2 + a q + 6 N H 3 ( a q ) N i N H 3 6 2 + ( M ) [ N i 2 + ] ( M ) [ N H 3 ] ( M ) Initial (I) 0.004 0 0.476 Change ( C) - x + x + x Equilibrium ( E) 0.004 - x x 0.476 + x We make the simplifying assumption that x is much smaller than 0.00400 M . Therefore, we can ignore the x term in the equilibrium value of N i N H 3 6 2 + and [ N H 3 ] and use the simplified value in the K d expression: K d = N i 2 + [ N H 3 ] 6 [ N i ( N H 3 ) 6 ] 2 + 1.82 × 10 - 9 = x 0.476 6 ( 0.004 ) x = N i 2 + = 6.3 × 10 - 10 M Conclusion: The concentration N i 2 + at equilibrium is calculated using the formation constant expression of N i N H 3 6 2 + complex.

The concentration of N i N O 3 2 in the given solution The concentration of N i 2 + ( a q ) in the given solution Solution: N i N O 3 2 = 4.00 × 10 - 3 M N i 2 + = 6.3 × 10 - 10 M Explanation: Concept: Initial concentration of N i N O 3 2 can be calculated using its given moles and the volume of the solution. After addition of the 0.500 M N H 3 solution into N i ( N O 3 ) 2 solution, there would be the formation of a [ N i ( N H 3 ) 6 ] 2 + complex. This complex would be in the equilibrium with N i 2 + , so by using the formation constant equation, we can find out the concentration of N i 2 + at equilibrium. Formula: m o l a r i t y = m o l e s o f s o l u t e v o l u m e o f s o l u t i o n i n L K f = [ N i ( N H 3 ) 6 ] 2 + N i 2 + [ N H 3 ] 6 Given: M o l e s o f N i N O 3 2 = 1 m m o l × 1 m o l 1000 m m o l = 0.001 m o l v o l u m e o f s o l u t i o n = 250 m L × 1 L 1000 m L = 0.250 L N H 3 = 0.500 M Calculations: a) Calculation for the concentration of N i N O 3 2 in the solution: m o l a r i t y = m o l e s o f s o l u t e v o l u m e o f s o l u t i o n i n L m o l a r i t y = 0.001 m o l 0.250 L = 0.004 M = 4.00 × 10 - 3 M b) Calculation for the concentration of N i 2 + ( a q ) in the solution: The complex formation reaction is N i 2 + a q + 6 N H 3 a q ⇌ N i N H 3 6 2 + ( a q ) Using reaction stoichiometry, we calculate the consumed mole of N H 3 . 1 m o l N i ( N O 3 ) 2 reacts with 6 m o l N H 3 , so 0.00100 m o l N i ( N O 3 ) 2 will react with 0.00100 m o l N i ( N O 3 ) 2 × 6 m o l N H 3 1 m o l N i ( N O 3 ) 2 = 6 × 10 - 3 m o l N H 3 We can see that N H 3 is an excess reagent, so the remaining moles of ammonia after the reaction is 0.500 M N H 3 - 6 × 10 - 3 M N H 3 = 0.476 M N H 3 Since K f is large, we can assume that essentially all the N i 2 + ions are converted to complex ions and N i N H 3 6 2 + = 0.00400 M . A negligible concentration of free N i 2 + ions remains at equilibrium which is assumed as x . We can calculate the [ N i 2 + ] by considering how much N i ( N H 3 ) 6 dissociates: N i N H 3 6 2 + ⇌ N i 2 + a q + 6 N H 3 ( a q ) The equilibrium constant, K d for this reaction is the reciprocal of K f . K d = 1 K f = N i 2 + [ N H 3 ] 6 [ N i ( N H 3 ) 6 ] 2 + = 1 5.5 × 10 8 = 1.82 × 10 - 9 We can construct a RICE table incorporating the concentrations and concentration changes of the products and reactants. Reaction N i N H 3 6 2 + ( a q ) ⇌ N i 2 + a q + 6 N H 3 ( a q ) N i N H 3 6 2 + ( M ) [ N i 2 + ] ( M ) [ N H 3 ] ( M ) Initial (I) 0.004 0 0.476 Change ( C) - x + x + x Equilibrium ( E) 0.004 - x x 0.476 + x We make the simplifying assumption that x is much smaller than 0.00400 M . Therefore, we can ignore the x term in the equilibrium value of N i N H 3 6 2 + and [ N H 3 ] and use the simplified value in the K d expression: K d = N i 2 + [ N H 3 ] 6 [ N i ( N H 3 ) 6 ] 2 + 1.82 × 10 - 9 = x 0.476 6 ( 0.004 ) x = N i 2 + = 6.3 × 10 - 10 M Conclusion: The concentration N i 2 + at equilibrium is calculated using the formation constant expression of N i N H 3 6 2 + complex.

Solution Summary: The author explains how the formation constant equation can be used to find the concentration of Ni2+ ions in the solution at equilibrium.

Chemistry: An Atoms-Focused Approach (Second Edition)

Chemistry: An Atoms-Focused Approach (Second Edition)

2nd Edition

ISBN: 9780393614053

Author: Thomas R. Gilbert, Rein V. Kirss, Stacey Lowery Bretz, Natalie Foster

Publisher: W. W. Norton & Company

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Chemistry: An Atoms-Focused Approach (Second Edition)

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