Chem 162 Worksheet 7 with answers PDF

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Chemistry 162 Worksheet 7

Name __________ ID ___________

pH = pKA + log ([A-]/[HA])

Useful equations: Kp = Kc(RT)Δn Kw = 1 x 10-14

R = 0.08206atm∙L/mol∙K

1.) Which of the following is a Br∅nsted-Lowry acid? A) NH4+ B) CCl4 C) NH2D) NH3 E) Br2 + Bronsted-Lowry acid needs to have a H to donate upon in solution, D) NH3 is actually a weak base 2.) ________ is found in carbonated beverages due to the reaction of carbon dioxide with water. A) CH3COOH B) H2CO3 C) HCOOH D) C6H5COOH E) CH3CH2COOH Carbon dioxide and water make what? CO2 + H2O  H2CO3 3.) Which of the following solutions would have the highest pH? Assume that they are all 0.10 M in acid at 25∘C. The acid is followed by its Ka value. -4

A) HF, 3.5 × 10 -10 B) HCN, 4.9 × 10 -4 C) HNO2, 4.6 × 10

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D) HCHO2, 1.8 × 10 Highest pH means the least acidic or more basic. If they are all the same concentration then the only issue is the % dissociation which is expressed by the Ka. The smaller the Ka the less dissociation the less H+ formed then the most basic. HX + H2O  H+ + X-4

4.) Calculate the concentration of OH⁻ in a solution that contains 3.9 x 10 M H3O⁺ at 25°C. Identify the solution as acidic, basic, or neutral. -11 2.6 × 10 M, acidic + Kw = [H ][OH-]  insert [H+] = 3.9 × 10-4 M, [OH-] = 2.6 × 10-11 M 5.) Determine the pH of a 0.023 M HNO3 solution. A) 12.36 B) 3.68 C) 1.64 D) 2.30 E) 2.49 No Ka given so it assumed to be a strong acid (which it is) so 100% dissociation, thus the H+ = 0.023 M, take the –log [0.023] = 1.64 -9

6.) Determine the pOH of a 0.227 M C5H5N solution at 25°C. The Kb of C5H5N is 1.7 × 10 . A) 4.59 B) 9.41 C) 4.71 D) 10.14 E) 9.29 C5H5N + H2O  OH + HC5H5N 0.227 M

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K is really small so the assumption will work K = x2/0.227-x  x2/0.227  1.7x10-9 = x2/0.227  1.944x10-5 -log[1.944x10-5] = 4.70675 -10

7.) The acid-dissociation constant of hydrocyanic acid (HCN) at 25.0°C is 4.9 × 10 . What is the pH of an aqueous solution of 0.080 M sodium cyanide (NaCN)? -3 A) 11.11 B) 2.89 C) 1.3 × 10 D) 7.8 E) 13.9 CN-+ H2O  OH- + HCN 0.080 M

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K comes from the relation of Kw = Ka x Kb 1x10-14 = 4.9x10-10 x Kb, Kb =2.0408x10-5 K = x2/0.080-x  x2/0.080  2.0408x10-9 = x2/0.080  0.001277 -log[0.001277] = 2.89355 = pOH. 14 = pH + pOH  11.106 ~11.1...