Use the References to access important values if needed for this question (1) Wr
ID: 704159 • Letter: U
Question
Use the References to access important values if needed for this question (1) Write a net ionic equation for the reaction between CHCOO (aq) and HS (aq) that shows CH,CO0 (aq) behaving as a Bronsted-Lowry base. Use CH CO0H for acetic acid and CHjCOO for the acetate ion BL base BL acid BL acid BL base (2) Using the table of relative strengths of acids and bases on the information page, decide which would be favored at equilibrium for this reaction, reactants or products? Submit Answer Retry Entire Group 3 more group attempts remaining Previous NextExplanation / Answer
Dear student, posting these many numbers of questions in a single post goes against Chegg policy and we are instructed to answer only the first question in such situations.
Nevertheless, I have answered all your questions but please keep the above in mind in future. The solution of the 1st and 2nd image has been skipped due to lack of relative strength table in the question.
The solution of all other questions are as follows :
1. [OH-] = 10-3 M
[H+]*[OH-] = 10-14
[H+]*10-3 = 10-14
[H+] = 10-11 M
Since, [H+]<[OH-]
The solution is basic.
2. [OH-] = 10-2 M
[H+]*[OH-] = 10-14
[H+]*10-2 = 10-14
[H+] = 10-12 M
Since, [H+]<[OH-]
The solution is basic.
3. Let, the concentration of Hydroxide ions = [OH-]
And, the concentration of Hydrogen ions = [H+]
Now, [H+]*[OH-] = 10-14
[H+]*10-12= 10-14
[H+] = 10-2 M
pH = -log[H+]
pH = -log[10-2]
pH = 2
pH + pOH = 14
2 + pOH = 14
pOH = 12
4. pH + pOH = 14
pH + 10 = 14
pH = 4
pH = -log[H+]
[H+] = 10-pH
[H+] = 10-4
[OH-] = 10-pOH
[OH-] = 10-10
5. pH = 11
[H+] = 10-8
pH = -log[H+]
pH = -log[10 -8]
pH = 8
[OH-] = 10-9
[H+]*[OH-] = 10-1
[H+]*10-9 = 10-14
[H+] = 10-5
pH = -log[H+]
pH = -log[10-5]
pH = 5
Therefore, order of acidity :
Solution 1 : 1 ( least acidic)
Solution 2 : 2
Solution 3 : 3 (most acidic)
6. pH = -log[H+]
pH = -log[10 -11]
pH = 11
pOH = 7
pH = 14 - pOH
pH = 7
[OH-] = 10-13
[H+]*[OH-] = 10-14
[H+]*10-13= 10-14
[H+] = 10-1
pH = -log[H+]
pH = -log[10 -1]
pH = 1
Therefore, order of acidity :
Solution 1 : 1 ( least acidic)
Solution 2 : 2
Solution 3 : 3 (most acidic)
7. [OH-] = 6.69*10-2 M
[H+]*[OH-] = 10-14
[H+]*6.69*10-2= 10-14
[H+] = 1.49*10-13
pOH = -log[OH-]
pOH = -log[6.69*10-2]
pOH = 1.17
pH = 14 - pOH
pH = 14 - 1.17
pH = 12.83
8. pH = 9.70
pH + pOH = 14
pOH = 4.3
[OH-] = 10-pOH
[OH-] = 10-4.3
[H+] = 10-pH
[H+] = 10-9.7
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