HW-Solns-Chap-8-3530

HW-Solns-Chap-8-3530 - Chapter 8: Homework Solutions 8.11...

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Chapter 8: Homework Solutions 8.11 (a) n(H + ) = 0.144 mol/L x 0.025 L = 3.60x10 -3 mol n(OH - ) = 0.125 mol/L x 0.025 L = 3.12x10 -3 mol After neutralization, excess n(H + ) = 3.6x10 -3 mol - 3.12x10 -3 mol = 4.8x10 -4 mol M x L mol x V H n H 3 4 10 6 . 9 ) 025 . 0 025 . 0 ( 10 8 . 4 ) ( ] [ pH = -log([H + ]) = -log(9.6x10 -3 M) = 2.02 (b) n(H + ) = 0.15 mol/L x 0.025 L = 3.75x10 -3 mol n(OH - ) = 0.15 mol/L x 0.035 L = 5.25x10 -3 mol After neutralization, excess n(OH - ) = 5.25x10 -3 mol - 3.75x10 -3 mol = 1.50x10 -3 mol M x L mol x V OH n OH 2 3 10 5 . 2 ) 035 . 0 025 . 0 ( 10 50 . 1 ) ( ] [ pOH = -log([OH - ]) = -log(2.5x10 -2 M) = 1.60 pH = 14.0 - pOH = 14.0 - 1.60 = 12.40 (c) n(H + ) = 0.22 mol/L x 0.0212 L = 4.66x10 -3 mol n(OH - ) = 0.30 mol/L x 0.010 L = 3.00x10 -3 mol After neutralization, excess n(H + ) = 4.66x10 -3 mol - 3.00x10 -3 mol = 1.66x10 -3 mol M x L mol x V H n H 2 3 10 32 . 5 ) 010 . 0 0212 . 0 ( 10 66 . 1 ) ( ] [ pH = -log([H + ]) = -log(5.32x10 -2 M) = 1.27 8.12 The general rule is: (1) salt of strong acid and strong base is neutral. (2) salt of strong acid and weak base is acidic. (3) salt of weak acid and strong base is basic. (4) salt of weak acid and weak base is often close to neutral. (a) acidic; NH 4 + H + + NH 3 (b) basic; CO 3 2- + H 2 O HCO 3 - + OH - (c) basic; F - + H 2 O HF + OH - (d) neutral: Because K + is the conjugate acid of a strong base and Br - is the conjugate base of a strong acid.
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8.13 (a) Calculate the molar concentration of KC 2 H 3 O 2 (8.4 g) 1 mol 98.15 g 1 0.250 L 0.342 mol L 1 H 2 O(l) C 2 H 3 O 2 (aq) É HC 2 H 3 O 2 OH (aq) Initial 0.342 0 0 Change x + x + x Equilibrium 0.342 x 0.342 X x 342 . 0 342 . 0
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This note was uploaded on 04/12/2011 for the course CHEM 3520 taught by Professor Staff during the Fall '08 term at North Texas.

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HW-Solns-Chap-8-3530 - Chapter 8: Homework Solutions 8.11...

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