logarithms, ph, poh, pk a. logarithms logarithmpower to which you must raise a base number to obtain...
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Logarithms, pH, pOH, pKa
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LOGARITHMSLOGARITHM power to which you must raise a
base number to obtain the desired number
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Log (base 10)
• Logs (base 10) = exponent to which 10 must be raised to obtain desired number
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Log (base 10)
• Logs (base 10) = exponent to which 10 must be raised to obtain desired number
• log 1 = 0 because 1 = 100
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Log (base 10)
• Logs (base 10) = exponent to which 10 must be raised to obtain desired number
• log 1 = 0 because 1 = 100
• Common log of a number that is a power of 10 is always a whole number
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Log (base 10)
•log is exponent of 10 WITH ITS SIGN eg -4 log 10-4 = -4
• Fractional power of 10 on calculator, enter number and use log button,
log (2 x 10-4) = -3.7
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Inverse log (base 10)
• Inverse log used to find number corresponding to log = 10x
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Inverse log (base 10)• Inverse log used to find number corresponding to log = 10x
• Whole integer log: use log (given number) with its sign as power of 10
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Inverse log (base 10)• Inverse log used to find number corresponding to log = 10x
• Whole integer log: use log (given number) with its sign as power of 10
• Decimal log: enter number WITH ITS SIGN Use the:
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Inverse log (base 10)• Inverse log used to find number corresponding to log = 10x
• Whole integer log: use log (given number) with its sign as power of 10
• Decimal log: enter number WITH ITS SIGN and use either
• 10x button OR
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Inverse log (base 10)• Inverse log used to find number corresponding to log = 10x
• Whole integer log: use log (given number) with its sign as power of 10
• Decimal log: enter number WITH ITS SIGN and use either
• 10x button OR
• [2nd][log] buttons
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Example #1 – inverse logWhat number corresponds to a
log= -5.34?
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Example #1 – inverse logWhat number corresponds to a
log= -5.34?
Inverse log –5.34 = 10-5.34
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Example #1 Solution
[2nd]log –5.34 = 10-5.34 = 4.6 x 10-6
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pH
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pH
pH A measure of the acidity of aqueous solutions
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pH
pH A measure of the acidity of aqueous solutions
pH = - log [H3O+]
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pH
pH A measure of the acidity of aqueous solutions
pH = - log [H3O+]
pOH = - log [OH-]
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pH
pH A measure of the acidity of aqueous solutions
pH = - log [H3O+]
pOH = - log [OH-]
pK = - log [Keq]
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pH
pH A measure of the acidity of aqueous solutions
pH = - log [H3O+]
pOH = - log [OH-]
pK = - log [Keq]
p(anything) = - log (anything)
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[H3O+] from pH[OH-] from pOH
[H3O+] = 10 - pH
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[H3O+] from pH[OH-] from pOH
[H3O+] = 10 - pH
[OH-] = 10 - pOH
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[H3O+] from pH[OH-] from pOH
[H3O+] = 10 - pH
[OH-] = 10 - pOH
• the higher the pH, the lower the [H3O+] in the solution
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pH in a NEUTRAL SOLUTION
neutral solution: [H3O+] = [OH-] = 1.0 x 10-7 M
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pH in a NEUTRAL SOLUTION
neutral solution: [H3O+] = [OH-] = 1.0 x 10-7 M
pH = pOH = 7.00
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SUMMARY OF pH
pH pOH
acidic < 7
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SUMMARY OF pH
pH pOH
acidic < 7
neutral 7
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SUMMARY OF pH
pH pOH
acidic < 7
neutral 7
basic > 7
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SUMMARY OF pH
pH pOH
acidic < 7 > 7
neutral 7
basic > 7
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SUMMARY OF pH
pH pOH
acidic < 7 > 7
neutral 7 7
basic > 7
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SUMMARY OF pH
pH pOH
acidic < 7 > 7
neutral 7 7
basic > 7 < 7
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pH CHANGE
• change of 1 pH unit means [H3O+] has undergone a 10-fold change
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pH CHANGE
• change of 1 pH unit means [H3O+] has undergone a 10-fold change
• pH 5 vs pH 6:
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pH CHANGE
• change of 1 pH unit means [H3O+] has undergone a 10-fold change
• pH 5 vs pH 6:
• change of one pH unit; change [ ] by factor of 10
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pH CHANGE
• change of 1 pH unit means [H3O+] has undergone a 10-fold change
• pH 5 vs pH 6:
• change of one pH unit; change [ ] by factor of 10
• pH 5 10X more acidic than pH 6
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pH CHANGE
• change of 1 pH unit means [H3O+] has undergone a 10-fold change
• pH 5 vs pH 6:
• change of one pH unit; change [ ] by factor of 10
• pH 5 10X more acidic than pH 6
• pH 4 100X more acidic than pH 6
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pKW
Kw = [H3O+][OH-] = 1.0 x 10-14
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pKW
Kw = [H3O+][OH-] = 1.0 x 10-14
pH + pOH = 14.00 = pKw
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pH of STRONG ACID
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Example #2 Calculations Involving pH and pOH
Calculate [H3O+], pH, [OH-], and pOH for a 0.015 M HNO3 solution.
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Example #2 Solution
HNO3(aq) + H2O H3O+(aq) + NO3-(aq)
0.015 M 0.015 M
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Example #2 Solution
HNO3(aq) + H2O H3O+(aq) + NO3-(aq)
0.015 M 0.015 M
pH = - log [H3O+]
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Example #2 Solution
HNO3(aq) + H2O H3O+(aq) + NO3-(aq)
0.015 M 0.015 M
pH = - log [H3O+] = - log 0.015 = - (-1.82) = 1.82
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Example #2 Solution
pH + pOH = 14.00
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Example #2 Solution
pH + pOH = 14.00
pOH = 14.00 - pH
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Example #2 Solution
pH + pOH = 14.00
pOH = 14.00 - pH = 14.00 - 1.82 = 12.18
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Example #2 Solution
pH + pOH = 14.00
pOH = 14.00 - pH = 14.00 - 1.82 = 12.18
pOH = - log [OH-]
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Example #2 Solution
pH + pOH = 14.00
pOH = 14.00 - pH = 14.00 - 1.82 = 12.18
pOH = - log [OH-]
12.18 = - log [OH-]
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Example #2 Solution
pH + pOH = 14.00
pOH = 14.00 - pH = 14.00 - 1.82 = 12.18
pOH = - log [OH-]
12.18 = - log [OH-]
[OH-] = [2nd ]log -12.18
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Example #2 Solution
pH + pOH = 14.00
pOH = 14.00 - pH = 14.00 - 1.82 = 12.18
pOH = - log [OH-]
12.18 = - log [OH-]
[OH-] = [2nd] log -12.18 = 10-12.18
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Example #2 Solution
pH + pOH = 14.00
pOH = 14.00 - pH = 14.00 - 1.82 = 12.18
pOH = - log [OH-]
12.18 = - log [OH-]
[OH-] = [2nd]log -12.18 =10-12.18 = 6.7 x 10-13 M
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pH of STRONG BASE
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Example #3 Calculations Involving pH and pOH
Calculate [H3O+], pH, [OH-], and pOH for a 0.015 M Ca(OH)2 solution.
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Example #3 Solution
Ca(OH)2(aq) Ca2+(aq) + 2 OH-(aq)
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Example #3 Solution
Ca(OH)2(aq) Ca2+(aq) + 2 OH-(aq) 0.015 M 2 x 0.015 M = 0.030 M
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Example #3 Solution
Ca(OH)2(aq) Ca2+(aq) + 2 OH-(aq) 0.015 M 2 x 0.015 M = 0.030 M
pOH = - log [OH-]
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Example #3 Solution
Ca(OH)2(aq) Ca2+(aq) + 2 OH-(aq) 0.015 M 2 x 0.015 M = 0.030 M
pOH = - log [OH-] = - log 0.030 = - (-1.52) = 1.52
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Example #3 Solution
pH + pOH = 14.00
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Example #3 Solution
pH + pOH = 14.00
pH = 14.00 - pOH
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Example #3 Solution
pH + pOH = 14.00
pH = 14.00 - pOH = 14.00 - 1.52 = 12.48
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Example #3 Solution
pH + pOH = 14.00
pH = 14.00 - pOH = 14.00 - 1.52 = 12.48
pH = - log [H3O+]
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Example #3 Solution
pH + pOH = 14.00
pH = 14.00 - pOH = 14.00 - 1.52 = 12.48
pH = - log [H3O+]
12.48 = - log [[H3O+]
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Example #3 Solution
pH + pOH = 14.00
pH = 14.00 - pOH = 14.00 - 1.52 = 12.48
pH = - log [H3O+]
12.48 = - log [[H3O+]
[H3O+] = [2nd]log -12.48
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Example #3 Solution
pH + pOH = 14.00
pH = 14.00 - pOH = 14.00 - 1.52 = 12.48
pH = - log [H3O+]
12.48 = - log [[H3O+]
[H3O+] = [2nd]log -12.48 = 10-12.48
![Page 65: Logarithms, pH, pOH, pK a. LOGARITHMS LOGARITHMpower to which you must raise a base number to obtain the desired number](https://reader030.vdocument.in/reader030/viewer/2022032600/56649dc35503460f94ab5ad3/html5/thumbnails/65.jpg)
Example #3 Solution
pH + pOH = 14.00
pH = 14.00 - pOH = 14.00 - 1.52 = 12.48
pH = - log [H3O+]
12.48 = - log [[H3O+]
[H3O+] = [2nd]log -12.48 =10-12.48 =3.3 x 10-13 M
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pKa
• pKa = - log Ka
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pKa
• pKa = - log Ka
•The lower the pKa, the stronger the acid
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pKa
• pKa = - log Ka
•The lower the pKa, the stronger the acid
• Used with weak acids• Strong acid have Ka approaching infinity
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REVIEW
• Auto-ionization of H2O: Kw = [H3O+][OH-]
• pH = - log [H3O+] = - log [H+]
• pOH = - log [OH-]
• pK = - log K