Acid Base Chemistry

Chem 102 Solutions to Review Problems for Acid/Base Chemistry 1. Glacial acetic acid, pure HC2H3O2 (FW = 60.0), has a concentration of 17.54 M. If 85.5 mL of glacial acetic acid are diluted to 250 mL, what is the acetic acid concentration? This is a dilution problem; use M1V1 = M2V2. (17.54 M)(85.5 mL) = M2 (250 mL) M2 = 6.00 M 2. If 26 mL of this diluted acetic acid (see Prob. 1) are further diluted to exactly 800 mL, the solution pH is 2.74. Calculate Ka for acetic acid. First, there is a dilution, followed by an equilibrium calculation involving a solution of a weak acid. For the dilution, (6.00 M)(26 mL) = M2 (800. mL) and M2 = 0.195 M For the weak acid solution, HC2H3O2 + H2O init 0.195 M equil 0.195-x X H3O+ + C2H3O2~10-7 M ~x 0 x where x = increase in [C2H3O2-]. Since the pH is known, the [H3O+], and hence x, is known. pH = 2.74 Y [H3O+] = 1.82 x 10-3 M = x Chem 102 Solutions to Review Problems for Acid/Base Chemistry 3. If 13.2 g NaC2H3O2 (FW = 82.0) are added to the 800 mL of solution in Problem 2, what is the resulting pH? The addition of C2H3O2- to a solution of HC2H3O2 creates a HC2H3O2 / C2H3O2buffer. initially, [HC2H3O2] = 0.195 M and mol C2H3O2- = 13.2 g/82.0 g/mol = 0.161 mol [C2H3O2-] = 0.161 mol/0.800 L = 0.201 M HC2H3O2 + H2O init 0.195 M equil ~0.195 X H3O+ + C2H3O2~10-7 M x x = 1.66 x 10-5 M = [H3O+] 0.201 M ~0.201 Y pH = 4.78 Chem 102 Solutions to Review Problems for Acid/Base Chemistry 4. The resulting 800 mL of solution in Problem 3 is divided into two 400-mL samples. If 5.0 mL of 6.0 M HCl are added to one sample, and 5.0 mL of 6.0 M NaOH are added to the other, what is the resulting pH in each case? The added HCl is neutralized by the weak base and a new buffer is formed. mol HCl added = (6.0 mol/L)(0.0050 L) = 0.030 mol init mol C2H3O2- = (0.201 mol/L)(0.400 L) = 0.0804 mol init mol HC2H3O2 = (0.195 mol/L)(0.400 L) = 0.078 mol init mol mol after rxn H3O+ + C2H3O20.030 0.0804 0 0.0504 6 HC2H3O2 + H2O 0.078 0.108 [HC2H3O2] = 0.108 mol/0.405 L = 0.267 M [C2H3O2-] = 0.0504 mol/0.405 L = 0.124 M HC2H3O2 + H2O init 0.267 M equil ~0.267 X H3O+ + C2H3O2~10-7 M x x = 3.68 x 10-5 M = [H3O+] 0.124 M ~0.124 Y pH = 4.43 Chem 102 Solutions to Review Problems for Acid/Base Chemistry 4. (continued) The added NaOH is neutralized by the weak acid and a new buffer is formed. mol NaOH added = (6.0 mol/L)(0.0050 L) = 0.030 mol init mol HC2H3O2 = (0.195 mol/L)(0.400 L) = 0.078 mol init mol C2H3O2- = (0.201 mol/L)(0.400 L) = 0.0804 mol init mol mol after rxn OH- + HC2H3O2 0.030 0.078 0 0.048 6 C2H3O2- + H2O 0.0804 0.1104 [HC2H3O2] = 0.048 mol/0.405 L = 0.12 M [C2H3O2-] = 0.1104 mol/0.405 L = 0.273 M HC2H3O2 + H2O init 0.12 M equil ~0.12 X H3O+ + C2H3O2~10-7 M x x = 7.52 x 10-6 M = [H3O+] 5. 0.273 M ~0.273 Y pH = 5.12 Calculate the pH of 10-5 M HCl. This is a solution of a strong acid and [H3O+] ~ [acid]. [H3O+] = 10-5 M Y pH = 5.0 Chem 102 Solutions to Review Problems for Acid/Base Chemistry 6. If 0.050 mL of 6.0 M HCl is added to 400 mL of 10-5 M HCl, what is the resulting pH? The added HCl is combines with the initial HCl to form a new strong acid solution. mol H3O+ added = (6.0 mol/L)(0.000050 L) = 0.00030 mol init mol H3O+ = (10-5 mol/L)(0.400 L) = 0.000004 mol total mol H3O+ = 0.000304 mol [H3O+] = 0.000304 mol/0.40005 L = 7.6 x 10-4 M pH = 3.12 (Notice that the pH of this unbuffered solution decreased by almost 2 pH units with this small addition of acid . Adding 100 times this volume of acid to a buffer (Prob. 4) changed the buffer pH by only 0.34 unit.) Chem 102 Solutions to Review Problems for Acid/Base Chemistry 7. If 0.050 mL of 6.0 M NaOH is added to 400 mL of 10-5 M HCl, what is the resulting pH? The added NaOH reacts with the HCl. mol NaOH added = (6.0 mol/L)(0.000050 L) = 0.00030 mol init mol H3O+ = (10-5 mol/L)(0.400 L) = 0.000004 mol init mol mol after rxn OH- + H3O+ 6 2H2O 0.00030 0.000004 0.000296 0 The HCl is completely neutralized and you are left with a solution of a strong base. [OH-] = 0.000296 mol/0.40005 L = 7.4 x 10-4 M pOH = 3.13 Y pH = 10.87 (Notice that the pH of this unbuffered solution increased by almost 6 pH units with this small addition of base . Adding 100 times this volume of base to a buffer (Prob. 4) changed the buffer pH by only 0.35 unit.) Chem 102 Solutions to Review Problems for Acid/Base Chemistry 8. Characterize the solution formed (strong acid, strong, base, weak acid, weak base, buffer or neutral) when equal volumes of the following are mixed. Explain. 0.5 M NaOH + 0.5 M HC2H3O2 The following reaction occurs: HC2H3O2 + OH- 6 H2O + C2H3O2- Since the strong base exactly neutralizes the weak acid (same concentrations and same volumes), a solution containing a weak base (C2H3O2-) is formed. 0.5 M HCl + 0.2 M NaNO3 NaNO3 is a neutral electrolyte (NO3- is the conjugate base of a strong acid) so it will not react with the strong acid HCl. A strong acid solution is formed. 1.0 M HNO3 + 2.0 M K2CO3 The following reaction occurs: CO32- + H3O+ 6 H2O + HCO3- Only one-half of the CO32- is converted to HCO3- (the CO32- concentration is twice that of the acid), so a CO32-/HCO3- buffer solution is formed.