ions and molecules in solution from the largest concentration to the smallest

Ions and molecules in solution from the largest

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ions and molecules in solution from the largest concentration to the smallest is: H 2 O, CH 3 COO , Na + , CH 3 COOH, H 3 O + , OH (b) Adapt the methods used in the answers to Question 30. The equation for the equilibrium and the equilibrium expression are: CH 3 COOH + H 2 O( ) CH 3 COO (aq) + H 3 O + (aq) Κ α = [ΧΗ 3 ΧΟΟ - ][Η 3 Ο + ] [ΧΗ 3 ΧΟΟΗ] 4.95 γ ΝαΧΗ 3 ΧΟΟ 250. μΛ 1 μολ ΝαΧΗ 3 ΧΟΟ 82.0337 γ ΝαΧΗ 3 ΧΟΟ 1000 μΛ 1 Λ 1 μολ ΧΗ 3 ΧΟΟ - 1 μολ ΝαΧΗ 3 ΧΟΟ = 0.241 M CH 3 COO As the reactants decompose, the concentrations of the products increase stoichiometrically, until they reach equilibrium concentrations. CH 3 COOH CH 3 COO (aq) H 3 O + (aq) conc. init. (M) 0.150 0.241 0 change conc. (M) – x + x + x eq. conc. (M) 0.150 – x 0.241 + x x At equilibrium K a = (ξ29( 0.241+ ξ 29 (0.150- ξ29 = 1.8 × 10 –5 Assume x is very small and does not affect the sum or the difference. 233
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Chapter 17: Additional Aqueous Equilibria 1.8 × 10 –5 = (ξ29( 0.241 29 (0.150 29 x = 1.1 × 10 –5 M = [H 3 O + ] pH = – log[H 3 O + ] = – log(1.1 × 10 –5 ) = 4.95 (c) The strong base neutralizes the acid in the solution, CH 3 COOH. CH 3 COOH + OH (aq) CH 3 COO (aq) + H 2 O( ) (conc. OH ) = 80. μγ ΝαΟΗ 100. μΛ 1000 μΛ 1 Λ 1 γ 1000 μγ 1 μολ ΝαΟΗ 40.00 γ ΝαΟΗ 1 μολ ΟΗ - 1 μολ ΝαΟΗ = 0.020 M We take it as far to the products as possible, using the method of limiting reactants. CH 3 COOH OH (aq) CH 3 COO (aq) conc. init. (M) 0.150 0.020 0.241 change conc. (M) – 0.020 – 0.020 + 0.020 conc. final (M) 0.130 0 0.261 The solution is still a buffer solution, containing an acid-base conjugate pair, so we can find the pH using K a of the acid. CH 3 COOH + H 2 O( ) CH 3 COO (aq) + H 3 O + (aq) Κ α = [ΧΗ 3 ΧΟΟ - ][Η 3 Ο + ] [ΧΗ 3 ΧΟΟΗ] As the reactants decompose, the concentrations of the products increase stoichiometrically, until they reach equilibrium concentrations. CH 3 COOH CH 3 COO (aq) H 3 O + (aq) conc. init. (M) 0.130 0.261 0 change conc. (M) – x + x + x eq. conc. (M) 0.130 – x 0.261 + x x At equilibrium K a = (0.261+ ξ29( ξ 29 (0.130- ξ29 = 1.8 × 10 –5 Assume x is very small and does not affect the sum or the difference. 1.8 × 10 –5 = (0.261 29( ξ 29 (0.130 29 x = 9.0 × 10 –6 M = [H 3 O + ] pH = – log[H 3 O + ] = – log(9.0 × 10 –6 ) = 5.05 (d) CH 3 COOH(aq) + H 2 O( ) CH 3 COO (aq) + H 3 O + (aq) 77. Adapt the methods used in the answers to Questions 15 - 32. At equilibrium, pH = 9.00, [OH ] = 10 pH–14.00 = 10 9.00–14.00 = 1.0 × 10 –5 Κ β = [ΝΗ 4 + ][ΟΗ - ] [ΝΗ 3 ] 1.8 × 10 –5 = [ΝΗ 4 + ](1.0 10 -5 29 (0.93 29 [NH 4 + ] = 1.7 M ( Notice: This could also be calculated using the Henderson-Hasselbalch equation .) 234
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Chapter 17: Additional Aqueous Equilibria 400. μΛ 1 Λ 1000 μΛ 1.7 μολ ΝΗ 4 + 1 Λ 1 μολ ΝΗ 4 Χλ 1 μολ ΝΗ 4 + 53.49 γ ΝΗ 4 Χλ 1 μολ ΝΗ 4 Χλ = 36 g NH 4 Cl 78 . It is possible to adapt the methods used in the answers to Question 15 - 32; however, the Henderson- Hasselbalch equation is easily used when we know the pK a of the conjugate acid. pH = pK a + log [ ο -ετηψλβενζοατε ] [ ο -ετηψλβενζοιχαχιδ ] At equilibrium, pH = 4.0 and pK a = 3.79 pH – pK a = log [ ο -ετηψλβενζοατε ] [ ο -ετηψλβενζοιχαχιδ ] 4.0 – 3.79 = 0.2 = log [ ο -ετηψλβενζοατε ] [ ο
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