Lecture_overheads_-_Ch08

Lecture_overheads_-_Ch08 - CHEM 321 Quantitative Analysis...

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CHEM 321 – Quantitative Analysis Ch. 8 – Monoprotic Acid Base Equilibria 8-1 Strong Acids and Bases 8-2 Weak Acids and Bases 8-3 Weak Acid Equilibria 8-4 Weak Base Equilibria 8-5 Buffers
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[OH - ] = 0.10 M pOH = - log [OH - ] = 1.00 pH + pOH = 14.00 pH = 13.00 note 2 SF in mantissa 8-1 Strong Acids and Bases Strong acids and bases defined in Ch 6 (Table 6-2) Strong acids and bases dissociate completely (K → ) Example 1 Calculate pH of 0.10 M solution of KOH KOH K + + OH - Initial 0.10 0 0 Final 0 0.10 0.10
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Example 2 Calculate pH of 0.10 M solution of KOH ( using activities ) [K + ] = [OH - ] = 0.10 M (1:1 salt, μ = Μ29 ( 29 ( 29 69 . 0 16 . 0 10 . 0 1 51 . 0 log 51 . 0 log 2 2 = - = - = - = γ γ μ γ z ( 29 ( 29 [ ] [ ] ( 29 ( 29 ( 29 ( 29 84 . 12 10 4 . 1 log log pH 10 4 . 1 69 . 0 10 . 0 10 00 . 1 OH H 10 00 . 1 13 H 13 H H 14 OH OH H 14 = × - = - = × = = × = = × = - - - - + - + + + - + - + a a a a a K H w γ γ pH = 12.84 (using activities) pH = 13.00 (using concentrations)
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Example 3 Calculate pH of 1.0 × 10 - 8 M solution of KOH Reality check : Even a very dilute solution of a strong base should NOT have acidic pH! What’s going on here and what are we not considering? [OH - ] = 1.0 × 10 - 8 M pOH = 8.00 pH = 14.000 - pOH = 6.00 pH = 6.00 ???
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Example 4 Calculate pH of 1.0 × 10 - 8 M solution of KOH ( using the systematic approach ) charge balance: [K + ] + [H + ] = [OH - ] [1] mass balance: [K + ] = 1.0 × 10 - 8 M [2] equilibrium reaction: K w = [H + ] [OH - ] = 1.0 × 10 - 14 [3] 3 equations, 3 unknowns [H + ], [OH - ], [K + ] Note that [K + ] is fixed (see above) Hence must solve 2 equations for 2 unknowns Do so by substituting equations 1 and 2 into equation 3
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[H + ] [OH - ] = 1.0 × 10 - 14 [H + ] (1.0 × 10 - 8 + [H + ]) = 1.0 × 10 - 14 [H + ] (1.0 × 10 - 8 ) + [H + ] 2 = 1.0 × 10 - 14 1[H + ] 2 + (1.0 × 10 - 8 )[H + ] + (-1.0 × 10 - 14 ) = 0 a x 2 + b x + c = 0 Example 4 (cont’d) quadratic equation a ac b b x 2 4 2 - ± - =
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Solve the quadratic equation for x x = 9.6 × 10 - 8 M ← this is the answer x = - 1.1 × 10 - 7 M can’t have negative conc! [H + ] = 9.6 × 10 - 8 M pH = 7.02 using systematic approach pH = 6.00 not using systematic approach Example 4 (cont’d) Explanation When dealing with dilute solutions of a base or acid, we must take K w into account to accurately compute pH!
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Explanation: At high to low concentrations (>10 - 6 M), pH is controlled by strong acid or base which dissociates completely At low concentrations (10 - 6 to 10 - 8 M), pH is controlled by both strong acid or base and K w (must use systematic approach to solve) At very low concentrations (<10 - 8 M), pH is controlled by dissociation of water ( K w ) and is neutral (7.000)
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Example 5 Calculate pH of 10. M solution of HCl [H + ] = 10. M pH = -1.00 Note pH can be negative! Example 6 Calculate pH of 10. M solution of NaOH [OH - ] = 10. M pOH = -1.00 pH = 15.00 Note pH can be > 14!
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8-2 Weak Acids and Bases Review of weak acid/base equilibria from general chemistry K a ’s tabulated in Text ( Appendix G ), must compute K b ’s [ ][ ] [ ] HA A H a - + = K A - + H 2 O ↔ HA + OH - HA H + + A - [ ] [ ] [ ] - - = A OH HA b K K a K b = K w H 2
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