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Salts

Salts. neutralization reactions. acids. bases. strong base. non-hydrolyzing salt. pH. =. 7.0. strong acid +. strong base. hydrolyzing salt. pH 7.0. >. weak acid +. <. weak base. hydrolyzing salt. pH 7.0. strong acid +. ?. weak base. hydrolyzing salt. pH =.

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Salts

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  1. Salts neutralization reactions acids bases strong base non-hydrolyzing salt pH = 7.0 strong acid+ strong base hydrolyzing salt pH 7.0 > weak acid+ < weak base hydrolyzing salt pH 7.0 strong acid+ ? weak base hydrolyzing salt pH = weak acid+ CH3COOH NaOH 1.00 M 1.00 M 500 mL

  2. CH3COOH CH3COO- + H+  1.00 M Ka = 1.8 x 10-5 = x2 500 mL 1.00 - x [CH3COOH] [CH3COO-] [H+] I 1.00 0.00 0.00 C -x +x +x E 1.00 - x x x 4.24 x 10-3 x = = [H+] . pH = 2.37

  3. Equivalence Point CH3COOH NaOH 1.00 M mol acid = mol base 1.00 M 500 mL 1.00 mol L x 0.500 L = 0.500 mol acid 0.500 mol base  1.00 mol L = 0.500 L base  CH3COOH CH3COO- + H+ + OH- H2O + CH3COO-  0.500 mol [CH3COO-] = = 0.50 M 0.500 + 0.500 L CH3COO- + H2O  CH3COOH + OH- strong conjugate base

  4. . Equivalence Point CH3COOH NaOH 1.00 M 1.00 M 500 mL CH3COO- + H2O  CH3COOH + OH- = 1 x 10-14 = 5.56 x 10-10 Kb = Kw 1.8 x 10-5 Ka . x = [OH-] = 1.67 x 10-5 pH = 9.22 5.56 x 10-10 = [CH3COOH] [OH-] = x2 0.500 - x [CH3COO-]

  5. . . CH3COOH NaOH 1.00 M 0.100 mol H+ = 0.100 mol OH- 1.00 M 500 mL add 100 mL 0.500 mol - 0.100 mol = .400 mol CH3COOH = 0.667 M 0.600 L 0.100 mol = 0.167 M CH3COO- 0.600 L . CH3COOH  CH3COO- + H+ [CH3COOH] [CH3COO-] [H+] I 0.667 0.167 0.00 Ka= 1.8 x 10-5 = (0.167 + x) (x) C -x +x +x (0.667 – x) E 0.667 - x 0.167 + x x x = 7.20 x 10-5 pH = 4.14

  6. . . Common ion effect presence of conjugate base inhibits dissociation weak acid or base conjugate base or acid + . . CH3COOH  CH3COO- + H+ . [CH3COOH] [CH3COO-] [H+] I 0.667 0.00 0.00 Ka= 1.8 x 10-5 = (x) (x) C -x +x +x (0.667 – x) E 0.667 - x x x x = 3.46 x 10-3 pH = 2.46

  7. [A-] [HA] [A-] [HA] Henderson-Hasselbalch Equation  HA + A- H+ Ka = [H+] [A-] [H+] = Ka [HA] [A-] [HA] - log [H+] = - log Ka + log pH = pKa + log

  8. [A-] [HA] . . . half-way point CH3COOH NaOH 1.00 M 0.250 mol H+ = 0.250 mol OH- 1.00 M 500 mL add 250 mL 0.500 mol - 0.250 mol = .250 mol CH3COOH = 0.333 M 0.750 L 0.250 mol = 0.333 M CH3COO- 0.750 L . Henderson-Hasselbalch Equation pH = pKa + log pH = pKa pH = 4.74 + log .333 = 4.74 .333

  9. [A-] [HA] . . . . CH3COOH NaOH 1.00 M 0.375 mol H+ = 0.375 mol OH- 1.00 M 500 mL add 375 mL 0.500 mol - 0.375 mol = .125 mol CH3COOH = 0.143 M 0.875 L 0.375 mol = 0.429 M CH3COO- 0.875 L . Henderson-Hasselbalch Equation pH = pKa + log pH = 4.74 + log .429 = 5.22 .143

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