CONCENTRATION PRACTICE PROBLEMS
Problem 1:
[H+]= 0.1M pH = pOH= [OH-]= |
Problem Formulas:
pH= -log[H+] pOH=-log[OH-] pH+pOH=14 [H+]= 2nd log[-pH] [OH-]=2nd log(pOH) [H+] x [OH-] = 1x 10E-14 |
Problem 3:
[H+] = 0.01M
[OH-]=
pH=
pOH=
[H+] = 0.01M
[OH-]=
pH=
pOH=
Answers:
1. [OH-]= pH = -log[H+] = -log[0.1 ]= 1 pOH= ph+pOH=14 = 14-1= 13 [OH-]= 2nd log [-pOH] = 2nd log [-13] = 1E-13M 2. 1x10-8 M NaOH (strong base) [H+]= 1x10-6 or 1E-6 [OH-] = 1x10-8 M pH=6 pOH=-log[OH-] =8 3. [H+] = 0.01M[OH-]= 1E12 pH= 2 pOH= 12 |
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