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**Unformatted text preview: **Chemistry 14CL Acid-Base Chemistry Review Key 1. Barium hydroxide furnishes two moles of hydroxide in solution for each mole of barium hydroxide that you start with, and HCl donates one mole of protons/mole of acid. Thus the concentration of [OH-] = 0.0800 M. Since equivalents of acid = equivalents of base, (X mL) (0.0800 M) = (15.0 mL)(0.100 M) and X = 18.75 mL. 2. Sulfuric acid furnishes two moles of protons for each mole of acid. Thus the concentration of [H + ] = 0.500 M. Since moles of [H + ] = moles of [OH-], (X mL)(0.500 M) = (38.40 mL)(0.200 M) and X = 15.4 mL. 3. moles of HCl = moles of NaOH = 0.0125 moles Molecular weight of NaOH = 40.0g, Weight of NaOH required = 0.50 g 4. This is a strong base which dissociates completely in solution. [OH-] = 0.037M; pOH = 1.43; pH = 12.57 5. [H+] = 0.030, pH = 1.5 pOH = (14-1.5) = 12.5 [OH-] =3.3 x 10-13 6. (a) [H+] = 0.05 M pH = 1.3 (b) [H+] = 1.2 x 10-4 pH = 3.9 7. (a) pOH = 1.3, pH = 12.7, [H+] = 2 x 10-13 (b) pOH = 7, pH = 7, [H+] = 1 x 10- 7 8. (a) [H+] = 3.0 10-3 M, pH =2.5 (b) [H+] = 8.9 x 10-5 , pH = 4.0 9. This is a limiting reagent problem only. The total moles of nitric acid involved in the titration is 2.00 x 10-3 . Therefore, half way through the titration, there are 1.00 x 1-3 moles left untitrated. The volume of the solution at this point in the titration can be calculated from the original 10 mL of nitric acid and the added 10 mL of base solution giving a solution volume of 20 mL. Therefore, the [H + ] of the solution at this point in the titration is 1.00 x 10-3 μολ 0.020Λ = 5.0ξ10-2 Μ and the pH is 1.3 10. The calculations differs from #9 only in the fact that you have a different amount of unreacted nitric acid and solution volume. 1.60 x 10- 3 mol 0.014L = 0.114M and the theoretical pH is 0.94 11. This is also a limiting reagent problem. However, since it also involves a weak acid, the equilibrium expression must be used to calculate [H + ] after the stoichiometry is determined from the limiting reagent calculations as was done in the problem above. ...

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