Lecnotes22 - MIT OpenCourseWare http/ocw.mit.edu 5.111 Principles of Chemical Science Fall 2008 For information about citing these materials or our

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MIT OpenCourseWare http://ocw.mit.edu 5.111 Principles of Chemical Science Fall 2008 For information about citing these materials or our Terms of Use, visit: http://ocw.mit.edu/terms .
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22.1 5.111 Lecture Summary #22 Acid/Base Equilibrium Continued (Chapters 10 and 11) Topics: Equilibrium involving weak bases, pH of salt solutions, and buffers From Wednesday’s handout 2. Base in water NH 3 (aq) + H 2 O (l) NH 4 + (aq) + OH - (aq) [NH 4 + ] [OH - ] Base ionization constant K b = [NH 3 ] K b is 1.8 x 10 -5 at 25°C. This small value tells us that only a small amount of NH 3 is present as NH 4 + . A strong base reacts essentially completely to give OH - (aq) when put in water. NH 3 is not a strong base. It is a moderately weak base. B (aq) + H 2 O (l) BH + (aq) + OH - (aq) BASE (B) IN WATER A - (aq) + H 2 O (l) HA (aq) + OH - (aq) BASE (A - ) IN WATER pK b = -log K b larger K b , stronger base larger pK b , weaker base 3. Conjugate acids and bases The stronger the acid, the weaker its conjugate base. The stronger the base, the weaker its conjugate acid. Consider conjugate acid-base pair NH 3 and NH 4 + : NH 3 (aq) + H 2 O (l) NH 4 + (aq) + OH - (aq) NH 4 + (aq) + H 2 O (l) H 3 O + (aq) + NH 3 (aq) Multiply K's together and get: + [NH 3 ][H 3 O + ] x [NH 4 ][OH - ] [H 3 O + ][OH - ] K a x K b = = [NH 4 + ] [NH 3 ]
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K a x K b = K w log K a + log K b = log K w or pK a + pK b = pK w = 14.00 Strong acid HA (aq) + H 2 O (l) H 3 O + (aq) + A - (aq) Strong base B (aq) + H 2 O (l) BH + (aq) + OH - (aq) 4. Relative strengths of acids Is HNO 3 or NH 4 + a stronger acid? Will the reaction lie far to the right or left? - HNO 3 (aq) + NH 3 (aq) NO 3 (aq) + NH 4 + (aq) [NO 3 - ][NH 4 + ] K = [HNO 3 ][NH 3 ] consider each acid separately: - 1. HNO 3 (aq) + H 2 O (l) H 3 O + (aq) + NO 3 (aq) - [H 3 O + ][NO 3 ] K a (HNO 3 ) = = 20. [HNO 3 ] 2. NH 4 + (aq) + H 2 O (l) H 3 O + (aq) + NH 3 (aq) [H 3 O + ][NH 3 ] K a (NH 4 + ) = = 5.6 x 10 -10 [NH 4 + ] Subtract equation 2 from 1 and divide the corresponding equilibrium constants. - [H 3 O + ][NO 3 ] K (HNO 3 ) [HNO 3 ] [NO 3 - ][NH 4 + ] K = a = = = 20. = 3.6 x 10
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This note was uploaded on 11/27/2011 for the course CHEMICAL E 20.410j taught by Professor Rogerd.kamm during the Spring '03 term at MIT.

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Lecnotes22 - MIT OpenCourseWare http/ocw.mit.edu 5.111 Principles of Chemical Science Fall 2008 For information about citing these materials or our

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