Chem 162-2011 Lecture 12

Chem 162-2011 Lecture 12 - CHEMISTRY 162-2011 LECTURE 12...

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CHEMISTRY 162-2011 LECTURE 12 ANNOUNCEMENTS E-MAIL ATTENDANCE Sign in EXAMS QUIZ Chem 162-2011 Lecture 12 1
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PLAN FOR TODAY : CHAPTER 15 ACIDS AND BASES EQUILIBRIUM Buffers Indicators Chem 162-2011 Lecture 12 2
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SIX STRONG ACIDS ET: Discuss strong acids and strong bases and the strengths of their conjugate bases and acids; practice both. Very weak base (i.e., not a base; K a Acid spectator ion)** ~10 9 HI + H 2 O H 3 O + + I - K cb = ~10 -23 ~10 8 HBr + H 2 O H 3 O + + Br - K cb = ~10 -22 ~10 6 HCl + H 2 O H 3 O + + Cl - K cb = ~10 -20 H 2 SO 4 + H 2 O H 3 O + + HSO 4 - K cb = ~0 HNO 3 + H 2 O H 3 O + + NO 3 - K cb = ~0 HClO 4 + H 2 O H 3 O + + ClO 4 - K cb = ~0 (HClO 3 + H 2 O H 3 O + + ClO 3 - )* K cb = ~0 *Strong, but not common, acid. Generally not considered as a strong acid. **These conjugate bases are known as “nominal bases”, i.e., a base by definition, but effectively not a base. All other acids are weak acids, e.g., HAc (acetic acid), H 2 SO 3 , RNH 3 + , BF 3 K a = 1.8x10 -5 HAc + H 2 O H 3 O + + Ac - K cb =5.6x10 -10 K a = 6.2x10 -10 HCN + H 2 O H 3 O + + CN - K cb =1.6x10 -5 Note that the stronger the acid, the weaker the CB; the weaker the acid, the stronger the CB. Arrhenius acid: Anything that provides a proton, e.g., HA, H 2 SO 3 , RNH 3 + LB Acid: Anything tending to give up a proton, e.g., HA, H 2 SO 3 , RNH 3 + STRONG BASES (Most Group 1A and 2A hydroxides [not HOH]) e.g. Very weak acid (i.e., not an acid; Base spectator ion) LiOH OH - + Li + NaOH OH - + Na + Mg(OH) 2 2OH - + Mg 2+ Ba(OH) 2 2OH - + Ba 2+ All other bases are weak bases, e.g., RNH 2 , CO 3 2- Arrhenius Base: Anything that provides an OH- group, e.g., NaOH LB Base: Anything that tends to react with a proton, e.g., NaOH, RNH 2 , CO 3 2- Chem 162-2011 Lecture 12 3
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ET: First discuss equations, then reverse equations invert K, then Ka x Kb = 10 -14 Write the equil. equations and identify K’s for the following reactions: Weak acid + water (e.g., acetic acid + H 2 O) : HA + H 2 O H 3 O + + A - 1 Assume K a = ~1 x 10 -5 Conjugate base + water (e.g., sodium acetate + H 2 O) : A - + H 2 O HA + OH - 3 K cb = K w /K a = (1 x 10 -14 )/(1 x 10 -5 ) = 1 x 10 -9 Conjugate base + H 3 O + (e.g., sodium acetate + H 3 O + ) :A - + H 3 O + → HA + H 2 O 2 K = 1/K a = 1/(1 x 10 -5 ) = 1 x 10 5 (or bring to completion; then solve either using HH or reverse + K a ) Weak acid + strong base (e.g., acetic acid + sodium hydroxide) :HA + OH - → H 2 O + A - 4 K = 1/(K w /K a ) = K a /K w = (1 x 10 -5 )/(1 x 10 -14 ) = 1 x 10 9 (or bring to completion; then solve either using HH or reverse + K a ) Weak base + water (e.g., NH 3 + H 2 O) : RNH 2 + H 2 O RNH 3 + + OH - Assume K b = ~1 x 10 -5 Conjugate acid + water (e.g., sodium acetate + H 2 O) : RNH 3 + + H 2 O RNH 2 + H 3 O + K ca = K w /K b = (1 x 10 -14 )/(1 x 10 -5 ) = (1 x 10 -9 ) Conjugate acid + OH - (e.g., ammonium chloride + OH - ) : RNH 3 + + OH - → RNH 2 + H 2 O K = 1/K b = 1/(1 x 10 -5 ) = 1 x 10 5 (or bring to completion; then solve either using HH or reverse + K b ) Weak base + strong acid (e.g., NH 3 + H 3 O + ) : RNH 2 + H 3 O + → RNH 3 + + H 2 O K = 1/(K w /K b ) = K b /K w = (1 x 10 -5 )/(1 x 10 -14 ) = 1 x 10 9 (or bring to completion; then solve either using HH or reverse + K b ) Chem 162-2011 Lecture 12 4
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CHEMICAL, ACID AND BASE EQUILIBRIA pH [H + ] 14 Basic 1x10 -14 H 2 O + H 2 O H 3 O + + OH - K w = 1x10 -14 7 Neutral K w = [H + ][OH - ] = 1 x 10 -14 pH + pOH = 14 K w = K a x K cb = K ca x K b = 1 x 10 -14 * *a & b = acid and base; ca & cb = conjugate acid and conjugate base pK a + pK b = 14 0 Acidic 1x10 o
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