University:
Boston University
Course:
CAS CH 214 | Organic Chemistry with Qualitative Analysis
Academic year:

2022
Views:

240

Pages:

2
Author:

pubicswprv

1. (2 points) An equal volume of 0.0010 M solution of C- (conjugate base of HC) is mixed with the solution of the strong acid and the final pH is 7.0. What is the pH of a 0.002 M solution of HC at 25oC? Based on the information we can conclude that C- is a strong Base that reacts 100% that means it’s conjugate acid so week it is weaker than water hence it will not change pH of water. pH=7 2 points R or W 2. (2 points) At 25oC an aqueous solution of an acid HA has pH = 3. Circle all of the following that must be true. The acid is a strong acid. The acid is a weak acid. Ka > Kw [H3O+] > [OH−] The conjugate base is strong. Ka ≈ Kw Ka < Kw [H3O+] ≈ [OH−] [H3O+] < [OH−] [HA] > [A−] [HA] ≈ [A−] [HA] < [A−] [HA] > [H3O+] [HA] ≈ [H3O+] [HA] < [H3O+] 3. (4 points) The solution of 0.20M NaD has a pH of 10.3. What is the K for this equilibrium reaction? D-(aq) + H2O(l) ⇌ HD(aq) + OH-(aq); −pH=3.7 [OH−] =10-pOH=10-3.7=1.99x10-4 ; 2 x10-4 =1.99x10-7=2x10-7 4. (2 points) Calculate the pH of a solution that is made by dissolving 0.8 moles of barium hydroxide in 400 mL of water at 25oC. [Ba(OH)2]=0.8mol/0.4L=2M [OH-]=4M pOH= -0.6 pH=14.6 1. (2 points) An equal volume of 0.0020 M solution of A- (conjugate base of HA) is mixed with the solution of the strong acid and the final pH is 7.0. What is the pH of a 0.002 M solution of HA at 25oC? Based on the information we can conclude that C- is a strong Base that reacts 100% that means it’s conjugate acid so week it is weaker than water hence it will not change pH of water. pH=7 2. (2 points) At 25oC an aqueous solution of an acid HA has pH = 4. Circle all of the following that must be true. The acid is a strong acid. The acid is a weak acid. The conjugate base is strong. Ka > Kw Ka ≈ Kw Ka < Kw [H3O+] > [OH−] [H3O+] ≈ [OH−] [H3O+] < [OH−] [HA] > [A−] [HA] ≈ [A−] [HA] < [A−] [HA] > [H3O+] [HA] ≈ [H3O+] [HA] < [H3O+] 3. (4 points) The solution of 0.20M NaD has a pH of 10.52. What is the K for this equilibrium reaction? D-(aq) + H2O(l) ⇌ HD(aq) + OH-(aq); −pH=3.48 [OH−] =10 -pOH =10-3.48=3.31x10-4 or 3.3 x10-4 =5.48x10-7=5.5x10-7 4. (2 points) Calculate the pH of a solution that is made by dissolving 0.4 moles of barium hydroxide in 500 mL of water at 25oC. [Ba(OH)2]=0.4mol/0.5L=0.8M [OH-]=1.6M pOH= -0.2 pH=14.2