Since 1 C 2 H 5 NH 3 ion is generated for each C 2 H 5 NH 3 Br mol C 2 H 5 NH 3

# Since 1 c 2 h 5 nh 3 ion is generated for each c 2 h

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Since 1 C 2 H 5 NH 3 + ion is generated for each C 2 H 5 NH 3 Br, mol C 2 H 5 NH 3 + = Since K b (C 2 H 5 NH 2 ) = 5.6 x 10 –4 , . Since , then . Check: The units (none) are correct. The magnitude of the answer makes physical sense because pH should be greater than the p K a of the acid because there is more base than acid. (c) Given: 10.0 g CH 3 COOH and 10.0 g CH 3 COONa in 150.0 mL solution Find: pH Other: K a (CH 3 COOH) = 1.8 x 10 –5 Conceptual Plan: Identify acid and base components then mL L and g CH 3 COOH mol acid = CH 3 COOH base = CH 3 COO CH 3 COOH then mol CH 3 COOH, L mol L –1 CH 3 COOH and g CH 3 COONa mol CH 3 COONa then mol CH 3 COONa, L mol L –1 CH 3 COONa mol L –1 CH 3 COO then mol L –1 CH 3 COOH, CH 3 COONa ( aq ) Na + ( aq ) + CH 3 COO ( aq ) mol L –1 CH 3 COO pH. Solution: and then and then . Acid = CH 3 COOH, so [acid] = [CH 3 COOH] = and base = CH 3 COO . Since 1 CH 3 COO ion is generated for each CH 3 COONa, [CH 3 COO ] = CH 3 COO = [base]. Then . pH = p K a + log [base] [acid] = - log (1.8 x 10 - 5 ) + log 0.812 612 mol L - 1 1.11 019 mol L - 1 = 4.61 0.812 612 mol L - 1 1.11 019 mol L - 1 M = mol L = 0.121 892 mol CH 3 COONa 0.1500 L = 0.812 612 mol L - 1 CH 3 COONa 10.0 g CH 3 COONa x 1 mol CH 3 COONa 82.04 g CH 3 COONa = 0.121 892 mol CH 3 COONa M = mol L = 0.166 528 mol CH 3 COOH 0.1500 L = 1.11 019 mol L - 1 CH 3 COOH 10.0 g CH 3 COOH x 1 mol CH 3 COOH 60.05 g CH 3 COOH = 0.166 528 mol CH 3 COOH 150.0 mL x 1 L 1000 mL = 0.1500 L pH = p K a + log [base] [acid] : : M = mol L : : 1 mol CH 3 COONa 82.04 g CH 3 COONa M = mol L : : 1 mol CH 3 COOH 60.05 g CH 3 COOH 1 L 1000 mL : : pH = p K a + log mol base mol acid = 10.75 + log 0.0232 8676 mol 0.00873 0852 mol = 11.18 p K a = 14 - p K b = 14 - 3.25 = 10.75 14 = p K a + p K b p K b = - log K b = - log (5.6 x 10 - 4 ) = 3.25 0.00873 0852 mol. 0.0232 8676 100 g solution x 1.10 g C 2 H 5 NH 3 Br 100 g solution x 1 mol C 2 H 5 NH 3 Br 125.99 g C 2 H 5 NH 3 Br = 0.00873 0852 mol C 2 H 5 NH 3 Br 100 g solution x 1.05 g C 2 H 5 NH 2 100 g solution x 1 mol C 2 H 5 NH 2 45.09 g C 2 H 5 NH 2 = 0.0232 8676 mol C 2 H 5 NH 2 pH = p K a + log [base] [acid] 14 = p K a + p K b p K b = - log K b : : : : : 1 mol C 2 H 5 NH 3 Br 125.99 g C 2 H 5 NH 3 Br 1.10 g C 2 H 5 NH 3 Br 100 g solution 1 mol C 2 H 5 NH 2 45.09 g C 2 H 5 NH 2 1.05 g C 2 H 5 NH 2 100 g solution mol C 2 H 5 NH 3 Br : g C 2 H 5 NH 3 Br : mol C 2 H 5 NH 2 : g C 2 H 5 NH 2 : Chapter 16 Aqueous Ionic Equilibrium 339
Check: The units (none) are correct. The magnitude of the answer makes physical sense because pH should be less than the p K a of the acid because there is more acid than base. 16.43 (a) Given: 50.0 mL of 0.15 mol L –1 HCOOH and 75.0 mL of 0.13 mol L –1 HCOONa Find: pH Other: K a (HCOOH) = 1.8 x 10 –4 Conceptual Plan: Identify acid and base components then mL HCOOH, mL HCOONa total mL then acid = HCOOH base = HCOO total mL = mL HCOOH + mL HCOONa mL HCOOH, mol L –1 HCOOH, total mL buffer mol L –1 HCOOH and mLHCOONa, mol L –1 HCOONa, total mL buffer mol L –1 HCOONa buffer mol L –1 HCOO then HCOONa ( aq ) Na + ( aq ) + HCOO ( aq ) K a , mol L –1 HCOOH, mol L –1 HCOO pH. Solution: Total mL = mL HCOOH + mL HCOONa = 50.0 mL + 75.0 mL = 125.0 mL. Then since rearrange to solve for M 2 .