Number Log log Crooo) -3 log C1o).-a 1000 0.01 Ix 104 -4 log (o-02), -1-69 .02 300 log (300 ), 2.47 3.2 2. Complete the following table. [H*] [OH] pH pOH A,B,N? .01 2. 7.
![PRE-LAB PROBLEMS:
1. Calculate logarithms and antilogarithms. You should be able to do the first three without
a calculator.
Number
Log
1000
log Crooo) - 3
0.01
1og C103).-a
Ix 10 4
-4
log (o-02), - 1-69
log (300 )- 2.47
.02
300
3.2
2. Complete the following table.
[H']
[OH']
pH
РОН
A,B,N?
.01
2
7](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F81ac168b-024d-494f-99de-3f3384ea3f4e%2F0b2af313-67e5-4d8a-892f-e977bdc27bf1%2F6mm6djr_processed.jpeg&w=3840&q=75)
pH is the measure of acidity whereas pOH is the measure of basicity.
For a solution, if pH < 7 then solution is acidic in nature.
if pH = 7 then soution is neutral.
If pH > 7 then solution is basic in nature.
pH = -log [H+]
pOH = -log [OH-]
also there is relation between pH and pOH,
pH + pOH = 14
1.) In the table, [H+] = 0.1 , pH = -log [H+] = -log(0.1)
therefore, pH = 1 ( solution is acidic)
pOH = 14 - pH = 14 - 1 = 13
Now, pOH = -log [OH-] =13
therefore, [OH-] = 10-13
2.) In the table, [OH-] = 0.01 , pOH = -log [OH-] = -log(0.01)
therefore, pOH = 2
pH = 14 - pOH = 14 - 2 = 12
pH = 12 (solution is basic)
Now, pH = -log [H+] =12
therefore, [H+] = 10-12
3.) In the table, pH = 2 , As pH = -log [H+] = 2 (acidic solution)
therefore, [H+]= 10-2
pOH = 14 - pH = 14 - 2 = 12
Now, pOH = -log [OH-] =12
therefore, [OH-] = 10-12
4.) In the table, pOH = 7 , As pOH = -log [OH-] = 7
therefore, [OH-] = 10-7
pH = 14 - pOH = 14 - 7 = 7
pH = 7 (solution is neutral)
Now, pH = -log [H+] =7
therefore, [H+]= 10-7
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