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Logarithms, pH, pOH, pK a
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LOGARITHMS LOGARITHMpower 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 = 10 0
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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 = 10 0 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 = 10 x
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Inverse log (base 10) Inverse log used to find number corresponding to log = 10 x 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 = 10 x 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 = 10 x Whole integer log: use log (given number) with its sign as power of 10 Decimal log: enter number WITH ITS SIGN and use either 10 x button OR
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Inverse log (base 10) Inverse log used to find number corresponding to log = 10 x Whole integer log: use log (given number) with its sign as power of 10 Decimal log: enter number WITH ITS SIGN and use either 10 x button OR [2 nd ][log] buttons
![Inverse log (base 10) Inverse log used to find number corresponding to log = 10 x Whole integer log: use log (given number) with its sign as power of 10 Decimal log: enter number WITH ITS SIGN and use either 10 x button OR [2 nd ][log] buttons](http://images.slideplayer.com/20/6028761/slides/slide_11.jpg "Inverse log (base 10) Inverse log used to find number corresponding to log = 10 x Whole integer log: use log (given number) with its sign as power of 10 Decimal log: enter number WITH ITS SIGN and use either 10 x button OR [2 nd ][log] buttons")
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Example #1 – inverse log What number corresponds to a log= -5.34?
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Inverse log –5.34 = 10 -5.34
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Example #1 Solution [2 nd ]log –5.34 = 10 -5.34 = 4.6 x 10 -6
![Example #1 Solution [2 nd ]log –5.34 = = 4.6 x 10 -6](http://images.slideplayer.com/20/6028761/slides/slide_14.jpg "Example #1 Solution [2 nd ]log –5.34 = = 4.6 x 10 -6")
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pH
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A measure of the acidity of aqueous solutions
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pH A measure of the acidity of aqueous solutions pH= - log [H 3 O + ]
![pH A measure of the acidity of aqueous solutions pH= - log [H 3 O + ]](http://images.slideplayer.com/20/6028761/slides/slide_17.jpg "pH A measure of the acidity of aqueous solutions pH= - log [H 3 O + ]")
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pH A measure of the acidity of aqueous solutions pH= - log [H 3 O + ] pOH= - log [OH - ]
![pH A measure of the acidity of aqueous solutions pH= - log [H 3 O + ] pOH= - log [OH - ]](http://images.slideplayer.com/20/6028761/slides/slide_18.jpg "pH A measure of the acidity of aqueous solutions pH= - log [H 3 O + ] pOH= - log [OH - ]")
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pH A measure of the acidity of aqueous solutions pH= - log [H 3 O + ] pOH= - log [OH - ] pK= - log [K eq ]
![pH A measure of the acidity of aqueous solutions pH= - log [H 3 O + ] pOH= - log [OH - ] pK= - log [K eq ]](http://images.slideplayer.com/20/6028761/slides/slide_19.jpg "pH A measure of the acidity of aqueous solutions pH= - log [H 3 O + ] pOH= - log [OH - ] pK= - log [K eq ]")
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pH A measure of the acidity of aqueous solutions pH= - log [H 3 O + ] pOH= - log [OH - ] pK= - log [K eq ] p(anything) = - log (anything)
![pH A measure of the acidity of aqueous solutions pH= - log [H 3 O + ] pOH= - log [OH - ] pK= - log [K eq ] p(anything) = - log (anything)](http://images.slideplayer.com/20/6028761/slides/slide_20.jpg "pH A measure of the acidity of aqueous solutions pH= - log [H 3 O + ] pOH= - log [OH - ] pK= - log [K eq ] p(anything) = - log (anything)")
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[H 3 O + ] from pH [OH - ] from pOH [H 3 O + ]= 10 - pH
![[H 3 O + ] from pH [OH - ] from pOH [H 3 O + ]= 10 - pH](http://images.slideplayer.com/20/6028761/slides/slide_21.jpg "[H 3 O + ] from pH [OH - ] from pOH [H 3 O + ]= 10 - pH")
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[H 3 O + ] from pH [OH - ] from pOH [H 3 O + ]= 10 - pH [OH - ]= 10 - pOH
![[H 3 O + ] from pH [OH - ] from pOH [H 3 O + ]= 10 - pH [OH - ]= 10 - pOH](http://images.slideplayer.com/20/6028761/slides/slide_22.jpg "[H 3 O + ] from pH [OH - ] from pOH [H 3 O + ]= 10 - pH [OH - ]= 10 - pOH")
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[H 3 O + ] from pH [OH - ] from pOH [H 3 O + ]= 10 - pH [OH - ]= 10 - pOH the higher the pH, the lower the [H 3 O + ] in the solution
![[H 3 O + ] from pH [OH - ] from pOH [H 3 O + ]= 10 - pH [OH - ]= 10 - pOH the higher the pH, the lower the [H 3 O + ] in the solution](http://images.slideplayer.com/20/6028761/slides/slide_23.jpg "[H 3 O + ] from pH [OH - ] from pOH [H 3 O + ]= 10 - pH [OH - ]= 10 - pOH the higher the pH, the lower the [H 3 O + ] in the solution")
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pH in a NEUTRAL SOLUTION neutral solution: [H 3 O + ] = [OH - ] = 1.0 x 10 -7 M
![pH in a NEUTRAL SOLUTION neutral solution: [H 3 O + ] = [OH - ] = 1.0 x M](http://images.slideplayer.com/20/6028761/slides/slide_24.jpg "pH in a NEUTRAL SOLUTION neutral solution: [H 3 O + ] = [OH - ] = 1.0 x M")
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pH in a NEUTRAL SOLUTION neutral solution: [H 3 O + ] = [OH - ] = 1.0 x 10 -7 M pH = pOH = 7.00
![pH in a NEUTRAL SOLUTION neutral solution: [H 3 O + ] = [OH - ] = 1.0 x M pH = pOH = 7.00](http://images.slideplayer.com/20/6028761/slides/slide_25.jpg "pH in a NEUTRAL SOLUTION neutral solution: [H 3 O + ] = [OH - ] = 1.0 x M pH = pOH = 7.00")
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SUMMARY OF pH pHpOH acidic< 7
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SUMMARY OF pH pHpOH acidic< 7 neutral 7
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SUMMARY OF pH pHpOH acidic< 7 neutral 7 basic> 7
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SUMMARY OF pH pHpOH acidic 7 neutral 7 basic> 7
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SUMMARY OF pH pHpOH acidic 7 neutral 7 7 basic> 7
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SUMMARY OF pH pHpOH acidic 7 neutral 7 7 basic> 7< 7
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pH CHANGE change of 1 pH unit means [H 3 O + ] has undergone a 10-fold change
![pH CHANGE change of 1 pH unit means [H 3 O + ] has undergone a 10-fold change](http://images.slideplayer.com/20/6028761/slides/slide_32.jpg "pH CHANGE change of 1 pH unit means [H 3 O + ] has undergone a 10-fold change")
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pH CHANGE change of 1 pH unit means [H 3 O + ] has undergone a 10-fold change pH 5 vs pH 6:
![pH CHANGE change of 1 pH unit means [H 3 O + ] has undergone a 10-fold change pH 5 vs pH 6:](http://images.slideplayer.com/20/6028761/slides/slide_33.jpg "pH CHANGE change of 1 pH unit means [H 3 O + ] has undergone a 10-fold change pH 5 vs pH 6:")
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pH CHANGE change of 1 pH unit means [H 3 O + ] has undergone a 10-fold change pH 5 vs pH 6: change of one pH unit; change [ ] by factor of 10
![pH CHANGE change of 1 pH unit means [H 3 O + ] has undergone a 10-fold change pH 5 vs pH 6: change of one pH unit; change [ ] by factor of 10](http://images.slideplayer.com/20/6028761/slides/slide_34.jpg "pH CHANGE change of 1 pH unit means [H 3 O + ] 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 [H 3 O + ] has undergone a 10-fold change pH 5 vs pH 6: change of one pH unit; change [ ] by factor of 10 pH 510X more acidic than pH 6
![pH CHANGE change of 1 pH unit means [H 3 O + ] has undergone a 10-fold change pH 5 vs pH 6: change of one pH unit; change [ ] by factor of 10 pH 510X more acidic than pH 6](http://images.slideplayer.com/20/6028761/slides/slide_35.jpg "pH CHANGE change of 1 pH unit means [H 3 O + ] has undergone a 10-fold change pH 5 vs pH 6: change of one pH unit; change [ ] by factor of 10 pH 510X more acidic than pH 6")
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pH CHANGE change of 1 pH unit means [H 3 O + ] has undergone a 10-fold change pH 5 vs pH 6: change of one pH unit; change [ ] by factor of 10 pH 510X more acidic than pH 6 pH 4100X more acidic than pH 6
![pH CHANGE change of 1 pH unit means [H 3 O + ] has undergone a 10-fold change pH 5 vs pH 6: change of one pH unit; change [ ] by factor of 10 pH 510X more acidic than pH 6 pH 4100X more acidic than pH 6](http://images.slideplayer.com/20/6028761/slides/slide_36.jpg "pH CHANGE change of 1 pH unit means [H 3 O + ] has undergone a 10-fold change pH 5 vs pH 6: change of one pH unit; change [ ] by factor of 10 pH 510X more acidic than pH 6 pH 4100X more acidic than pH 6")
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pK W K w = [H 3 O + ][OH - ] = 1.0 x 10 -14
![pK W K w = [H 3 O + ][OH - ] = 1.0 x](http://images.slideplayer.com/20/6028761/slides/slide_37.jpg "pK W K w = [H 3 O + ][OH - ] = 1.0 x")
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pK W K w = [H 3 O + ][OH - ] = 1.0 x 10 -14 pH + pOH = 14.00 = pK w
![pK W K w = [H 3 O + ][OH - ] = 1.0 x pH + pOH = = pK w](http://images.slideplayer.com/20/6028761/slides/slide_38.jpg "pK W K w = [H 3 O + ][OH - ] = 1.0 x pH + pOH = = pK w")
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pH of STRONG ACID
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Example #2 Calculations Involving pH and pOH Calculate [H 3 O + ], pH, [OH - ], and pOH for a 0.015 M HNO 3 solution.
![Example #2 Calculations Involving pH and pOH Calculate [H 3 O + ], pH, [OH - ], and pOH for a M HNO 3 solution.](http://images.slideplayer.com/20/6028761/slides/slide_40.jpg "Example #2 Calculations Involving pH and pOH Calculate [H 3 O + ], pH, [OH - ], and pOH for a M HNO 3 solution.")
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Example #2 Solution HNO 3 (aq) + H 2 O H 3 O + (aq) + NO 3 - (aq) 0.015 M 0.015 M
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Example #2 Solution HNO 3 (aq) + H 2 O H 3 O + (aq) + NO 3 - (aq) 0.015 M 0.015 M pH = - log [H 3 O + ]
![Example #2 Solution HNO 3 (aq) + H 2 O H 3 O + (aq) + NO 3 - (aq) M M pH = - log [H 3 O + ]](http://images.slideplayer.com/20/6028761/slides/slide_42.jpg "Example #2 Solution HNO 3 (aq) + H 2 O H 3 O + (aq) + NO 3 - (aq) M M pH = - log [H 3 O + ]")
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Example #2 Solution HNO 3 (aq) + H 2 O H 3 O + (aq) + NO 3 - (aq) 0.015 M 0.015 M pH = - log [H 3 O + ] = - log 0.015 = - (-1.82) = 1.82
![Example #2 Solution HNO 3 (aq) + H 2 O H 3 O + (aq) + NO 3 - (aq) M M pH = - log [H 3 O + ] = - log = - (-1.82) = 1.82](http://images.slideplayer.com/20/6028761/slides/slide_43.jpg "Example #2 Solution HNO 3 (aq) + H 2 O H 3 O + (aq) + NO 3 - (aq) M M pH = - log [H 3 O + ] = - log = - (-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 - ]
![Example #2 Solution pH + pOH = pOH = pH = = pOH = - log [OH - ]](http://images.slideplayer.com/20/6028761/slides/slide_47.jpg "Example #2 Solution pH + pOH = pOH = pH = = 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 - ]
![Example #2 Solution pH + pOH = pOH = pH = = pOH = - log [OH - ] = - log [OH - ]](http://images.slideplayer.com/20/6028761/slides/slide_48.jpg "Example #2 Solution pH + pOH = pOH = pH = = pOH = - log [OH - ] = - 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 - ] = [2 nd ]log -12.18
![Example #2 Solution pH + pOH = pOH = pH = = pOH = - log [OH - ] = - log [OH - ] [OH - ] = [2 nd ]log](http://images.slideplayer.com/20/6028761/slides/slide_49.jpg "Example #2 Solution pH + pOH = pOH = pH = = pOH = - log [OH - ] = - log [OH - ] [OH - ] = [2 nd ]log")
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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 - ] = [ 2 nd ] log -12.18 = 10 -12.18
![Example #2 Solution pH + pOH = pOH = pH = = pOH = - log [OH - ] = - log [OH - ] [OH - ] = [ 2 nd ] log =](http://images.slideplayer.com/20/6028761/slides/slide_50.jpg "Example #2 Solution pH + pOH = pOH = pH = = pOH = - log [OH - ] = - log [OH - ] [OH - ] = [ 2 nd ] log =")
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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 - ] = [2 nd ]log -12.18 =10 -12.18 = 6.7 x 10 -13 M
![Example #2 Solution pH + pOH = pOH = pH = = pOH = - log [OH - ] = - log [OH - ] [OH - ] = [2 nd ]log = = 6.7 x M](http://images.slideplayer.com/20/6028761/slides/slide_51.jpg "Example #2 Solution pH + pOH = pOH = pH = = pOH = - log [OH - ] = - log [OH - ] [OH - ] = [2 nd ]log = = 6.7 x M")
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pH of STRONG BASE
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Example #3 Calculations Involving pH and pOH Calculate [H 3 O + ], pH, [OH - ], and pOH for a 0.015 M Ca(OH) 2 solution.
![Example #3 Calculations Involving pH and pOH Calculate [H 3 O + ], pH, [OH - ], and pOH for a M Ca(OH) 2 solution.](http://images.slideplayer.com/20/6028761/slides/slide_53.jpg "Example #3 Calculations Involving pH and pOH Calculate [H 3 O + ], pH, [OH - ], and pOH for a M Ca(OH) 2 solution.")
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Example #3 Solution Ca(OH) 2 (aq) Ca 2+ (aq) + 2 OH - (aq)
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Example #3 Solution Ca(OH) 2 (aq) Ca 2+ (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) Ca 2+ (aq) + 2 OH - (aq) 0.015 M 2 x 0.015 M = 0.030 M pOH = - log [OH - ]
![Example #3 Solution Ca(OH) 2 (aq) Ca 2+ (aq) + 2 OH - (aq) M 2 x M = M pOH = - log [OH - ]](http://images.slideplayer.com/20/6028761/slides/slide_56.jpg "Example #3 Solution Ca(OH) 2 (aq) Ca 2+ (aq) + 2 OH - (aq) M 2 x M = M pOH = - log [OH - ]")
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Example #3 Solution Ca(OH) 2 (aq) Ca 2+ (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
![Example #3 Solution Ca(OH) 2 (aq) Ca 2+ (aq) + 2 OH - (aq) M 2 x M = M pOH = - log [OH - ] = - log = - (-1.52) = 1.52](http://images.slideplayer.com/20/6028761/slides/slide_57.jpg "Example #3 Solution Ca(OH) 2 (aq) Ca 2+ (aq) + 2 OH - (aq) M 2 x M = M pOH = - log [OH - ] = - log = - (-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 [H 3 O + ]
![Example #3 Solution pH + pOH = pH = pOH = = pH = - log [H 3 O + ]](http://images.slideplayer.com/20/6028761/slides/slide_61.jpg "Example #3 Solution pH + pOH = pH = pOH = = pH = - log [H 3 O + ]")
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Example #3 Solution pH + pOH = 14.00 pH = 14.00 - pOH = 14.00 - 1.52 = 12.48 pH = - log [H 3 O + ] 12.48 = - log [[H 3 O + ]
![Example #3 Solution pH + pOH = pH = pOH = = pH = - log [H 3 O + ] = - log [[H 3 O + ]](http://images.slideplayer.com/20/6028761/slides/slide_62.jpg "Example #3 Solution pH + pOH = pH = pOH = = pH = - log [H 3 O + ] = - log [[H 3 O + ]")
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Example #3 Solution pH + pOH = 14.00 pH = 14.00 - pOH = 14.00 - 1.52 = 12.48 pH = - log [H 3 O + ] 12.48 = - log [[H 3 O + ] [H 3 O + ] = [2 nd ]log -12.48
![Example #3 Solution pH + pOH = pH = pOH = = pH = - log [H 3 O + ] = - log [[H 3 O + ] [H 3 O + ] = [2 nd ]log](http://images.slideplayer.com/20/6028761/slides/slide_63.jpg "Example #3 Solution pH + pOH = pH = pOH = = pH = - log [H 3 O + ] = - log [[H 3 O + ] [H 3 O + ] = [2 nd ]log")
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Example #3 Solution pH + pOH = 14.00 pH = 14.00 - pOH = 14.00 - 1.52 = 12.48 pH = - log [H 3 O + ] 12.48 = - log [[H 3 O + ] [H 3 O + ] = [2 nd ]log -12.48 = 10 -12.48
![Example #3 Solution pH + pOH = pH = pOH = = pH = - log [H 3 O + ] = - log [[H 3 O + ] [H 3 O + ] = [2 nd ]log =](http://images.slideplayer.com/20/6028761/slides/slide_64.jpg "Example #3 Solution pH + pOH = pH = pOH = = pH = - log [H 3 O + ] = - log [[H 3 O + ] [H 3 O + ] = [2 nd ]log =")
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Example #3 Solution pH + pOH = 14.00 pH = 14.00 - pOH = 14.00 - 1.52 = 12.48 pH = - log [H 3 O + ] 12.48 = - log [[H 3 O + ] [H 3 O + ] = [2 nd ]log -12.48 =10 -12.48 =3.3 x 10 -13 M
![Example #3 Solution pH + pOH = pH = pOH = = pH = - log [H 3 O + ] = - log [[H 3 O + ] [H 3 O + ] = [2 nd ]log = =3.3 x M](http://images.slideplayer.com/20/6028761/slides/slide_65.jpg "Example #3 Solution pH + pOH = pH = pOH = = pH = - log [H 3 O + ] = - log [[H 3 O + ] [H 3 O + ] = [2 nd ]log = =3.3 x M")
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pK a pK a = - log K a
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pK a pK a = - log K a The lower the pK a, the stronger the acid
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pK a pK a = - log K a The lower the pK a, the stronger the acid Used with weak acids Strong acid have K a approaching infinity
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REVIEW Auto-ionization of H 2 O: K w = [H 3 O + ][OH - ] pH = - log [H 3 O + ] = - log [H + ] pOH = - log [OH - ] pK = - log K
![REVIEW Auto-ionization of H 2 O: K w = [H 3 O + ][OH - ] pH = - log [H 3 O + ] = - log [H + ] pOH = - log [OH - ] pK = - log K](http://images.slideplayer.com/20/6028761/slides/slide_69.jpg "REVIEW Auto-ionization of H 2 O: K w = [H 3 O + ][OH - ] pH = - log [H 3 O + ] = - log [H + ] pOH = - log [OH - ] pK = - log K")
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