The greater the acid strength of HA the less the base strength of A 6 Formal

The greater the acid strength of ha the less the base

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The greater the acid strength of HA, the less the base strength of A - 6
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Formal Concentration Formal Concentration (F): total number of moles of a compound dissolved in a liter. For a weak acid, it is the total amount of HA placed in the solution regardless of how much has been converted to A - 7
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Formal concentration and weak acid 8 HA A - + H + HA A - H + Initial Concentration F 0 0 Final Concentration F-x x x Example: Consider the composition of a solution containing 0.10 moles of weak acid in 1.0 L of water. We can say F = 0.10 M even if we do not know how much of the conjugate base was actually formed. Since it is a weak acid, it is only partially dissociated. At equilibrium, the concentration of HA is less than the formal concentration. From the mass balance: F = 0.10 M = [HA] + [A - ] Let x = [A - ] F = [HA] + x [HA] = F - x pH depends on amount of x
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Calculate pH of Weak Acid 9 We set up and solve the equation where [H + ] = [A - ] = x , and [HA] = F - x . Solve for x by solving quadratic equation HA A - H + Initial Concentration F 0 0 Final Concentration F-x x x HA A - + H + 𝐾 a = H + A [HA] = 𝑥 𝑥 F − 𝑥 = 𝑥 2 F − 𝑥 Note: For reasonable concentrations of weak acids (of reasonable strength), [H + ] from dissociation of acid >> contributions from dissociation of water For every mole of HA dissociated to A - , one mole of H + is formed, thus [H + ] = [A - ] = x ,
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Problem 1: Typical Weak Acid Problem CH 3 COOH CH 3 COO - H 3 O + Initial Concentration F 0 0 Final Concentration F - x = 0.50 - x x x CH 3 COOH + H 2 O H 3 O + + CH 3 COO - ax 2 + bx + c = 0 x = −?± ? 2 −4?? 2? 10 Problem: Calculate the pH of a solution of acetic acid (weak acid) if its formal concentration is 0.50M given its K a is 1.75 x 10 -5 . K a = [H 3 O + ] [CH 3 COO ] [CH 3 COOH] = 1.75 x 10 -5 K a = (x)(x) 0.50 − x = 1.75 x 10 -5 x 2 + (1.75 x 10 -5 )(x) - 0.875 x 10 -5 = 0 x = 2.94 x 10 -3 M = [H + ] pH = -log[H + ] = -log(2.94 x 10 -3 ) = 2.53 Solution: F = 0.50 M = [HA] + [A - ] Let x = [A - ] F = [HA] + x [HA] = F x = 0.50 - x
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Approximation: Simplification of weak-acid problem 11 Can always solve quadratic equation…. However, can first try to neglect x in the denominator (i.e. neglect x in F - x) If x ≤ 1% of F (i.e. x is 100 times ≤ F), then approximation is valid. As a simple rule: If K < (10 -4 )(F), then F - x F (try to neglect x) If K ≥ ( 10 -4 )(F), solve using the quadratic equation to reduce error on calculation
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Problem 2: Weak Acid Dissociation - Simplification Problem: Calculate the pH of a 0.050 M trimethylammonium chloride solution. 12 Solution: The pK a for trimethylammonium ion is 9.799 (see APPENDIX B). Screen shot of ebook: VitalSource Technologies LLC
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13 (CH 3 ) 3 NHCl (CH 3 ) 3 NH + + Cl - Assume this chloride salt is completely dissociated. Cl - essentially has no acidic or basic properties. (CH 3 ) 3 NH + (CH 3 ) 3 N + H + pK a = -logK a K a = 10 -pKa = 10 -9.799 = 1.59x10 -10 Since K a < (10 -4 )(F): 1.59 x 10 -10 < (10 -4 )(0.050), try the assumption F - x ≅ 𝐹 (CH 3 ) 3 NH + (CH 3 ) 3 N H + Initial Concentration F 0 0 Final Concentration F x F = 0.050 x x Problem 2: Weak Acid Dissociation Simplification (cont’d)
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K a = [H + ] [(CH 3 ) 3 N] [(CH 3 ) 3 NH + ] = 1.59 x 10 -10 K a = (x)(x) 0.050 = 1.59 x 10 -10 x 2 - 7.95 x 10 -12 = 0 x = 7.95 𝑥 10 −12 = 2.8195 x 10 -6 M = [H + ] pH = -log[H + ] = -log(2.8195 x 10 -6 ) = 5.55 14 Determine if assumption is valid: Is 100(x) ≤ F?
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