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18. A 1.00 L solution contains 1.20 mol of weak acid HX. If 0.50 mol NaOH is added to this solution, the pH of the resultant solution is 5.20. Calculate Kafor acid HX. (a) 2.5 x 10-4(b) 5.0 x 10-5(c) 4.5 x 10-6(d) 6.0 x 10-4(e) 1.2 x 10-5HX + OH-→H2O + X-pH = 5.20 [H3O+] = 10-5.20= 6.31 x 10-6Kw= [H3O+][OH-] = 1 x 10-14[OH-] = (1 x 10-14)/(6.31 x 10-6) = 1.58 x 10-9HX + OH-→H2O + Initial 1.20 0.50 Change Equilibrium 1.58 x 10-9X-
Assume that reactions with strong bases go to completion. X-0 +X +X X-CHEM 162-2007 EXAM II + ANSWERS CHAPTER 15A – ACID & BASE EQUILIBRIA ACID AND BASE EQUILIBRIA CALCULATIONS 19. Ascorbic acid, H2C6H6O6is a diprotic acid with Ka1= 7.9 x 10-5and Ka2= 1.6 x 10-12. In an 0.50 M aqueous solution of ascorbic acid, which of the following species would be present in the lowest concentration? +6-O
6-0 +X +X 2-0 +X +X M
47SALT EQUILIBRIA CONCEPTS/CALCULATIONSChem 162-2010 Final exam Acids and Bases - Chapter 15A Acid-base properties of salts concepts 3. Which of the following are BASICsalts? W. NH4I X. BaCl2Y. KNO3Z. NaF A. Z only B. Xand Y C. Y and Z D. W only E. W and X NH4I →NH4++ I-I-is such a weak base that it doesn’t act basic. I- + H2O →NR But NH4+is a weak acid. NH4++ H2O ←→NH3+ H3O+Therefore, solution is acidic. BaCl2→Ba2++ 2Cl-Group I and II cations are not acidic. Ba2+ + H2O →NR The conjugate bases of strong acids are not basic. Cl- + H2