1 / 4

Practice Quiz on Aqueous Equilibria: Buffers and pH Calculations

This practice quiz tests your understanding of aqueous equilibria, focusing on buffer solutions and pH calculations. You will determine which combinations of solutions create a pH 7 buffer, calculate the resulting pH from given reactions involving HCN and OH-, and analyze the dissociation of HF to find the pH at equilibrium. Each question is designed to reinforce concepts in acid-base chemistry and equilibrium. Work through the scenarios to enhance your problem-solving skills in aqueous equilibria.

cissy
Download Presentation

Practice Quiz on Aqueous Equilibria: Buffers and pH Calculations

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. PRACTICE QUIZ KEY AQUEOUS EQUILIBRIA (20 points)

  2. 1. (2 pts) Which makes a pH 7 buffer? • a. 1 M NaOH and 1 M NaCl • b. 1 M HC2H3O2 and 1 MNaC2H3O2 • c. 1 M HCl and 1 M NaCl • d. 1 M NH4Cl and 1 M NH3

  3. 2. (10 pts) Calculate the pH. • 0.200 M x 0.0750 L = 0.0150 mol • Step 1: OH- + HCN W H2O + CN- • Init. 0.0150 0.0150 0 • RXN -0.0150 -0.0150 +0.0150 • Final 0 0 0.0150 mol • (Must be equal at final) • Change mol to M; 0.0150 mol / total volume • V of OH-: V = mol / M = 0.0150 mol / 0.300 M = 0.0500 L • Total volume = 50.0 mL OH- + 75.0 mL HCN = 125.0 mL = 0.125 L • 0.0150 mol CN- / 0.125 L = 0.120 M CN- • Step 2: CN- + H2O W HCN + OH- • 0.120 – x x x • Kb = Kw/Ka = 1.0 x 10-14/4.0 x 10-10 = 2.5 X 10-5 • Kb = [HCN][OH-]/[CN-] = x2 / 0.120 • x = 1.7 x 10-3 = [OH-]; pOH = 2.78; pH = 11.22

  4. 3. (8 pts) What is the pH? • HF W H+ + F- • At equil.: 0.800-x x 0.500+x • Assume “x” is small enough to ignore… • Ka = [H+][F-]/[HF] = 4.0 x 10-4 • 4.0 x 10-4 = x(0.500)/(0.800) • x = 6.4 x 10-4 = [H+] • pH = 3.19

More Related