Logarithms, pH, pOH, pK a

LOGARITHMS LOGARITHMpower to which you must raise a base number to obtain the desired number

Log (base 10) Logs (base 10) = exponent to which 10 must be raised to obtain desired number

Log (base 10) Logs (base 10) = exponent to which 10 must be raised to obtain desired number log 1 = 0 because 1 = 10 0

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

Log (base 10) log is exponent of 10 WITH ITS SIGN eg -4 log = -4 Fractional power of 10 on calculator, enter number and use log button, log (2 x 10 -4) = -3.7

Inverse log (base 10) Inverse log used to find number corresponding to log = 10 x

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

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:

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

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

Example #1 – inverse log What number corresponds to a log= -5.34?

Inverse log –5.34 =

Example #1 Solution [2 nd ]log –5.34 = = 4.6 x 10 -6

pH

A measure of the acidity of aqueous solutions

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 + ] pOH= - log [OH - ]

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 ] p(anything) = - log (anything)

[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 [OH - ]= 10 - pOH

[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

pH in a NEUTRAL SOLUTION neutral solution: [H 3 O + ] = [OH - ] = 1.0 x M

pH in a NEUTRAL SOLUTION neutral solution: [H 3 O + ] = [OH - ] = 1.0 x M pH = pOH = 7.00

SUMMARY OF pH pHpOH acidic< 7

SUMMARY OF pH pHpOH acidic< 7 neutral 7

SUMMARY OF pH pHpOH acidic< 7 neutral 7 basic> 7

SUMMARY OF pH pHpOH acidic 7 neutral 7 basic> 7

SUMMARY OF pH pHpOH acidic 7 neutral 7 7 basic> 7

SUMMARY OF pH pHpOH acidic 7 neutral 7 7 basic> 7< 7

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 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: 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 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 pH 4100X more acidic than pH 6

pK W K w = [H 3 O + ][OH - ] = 1.0 x

pK W K w = [H 3 O + ][OH - ] = 1.0 x pH + pOH = = pK w

pH of STRONG ACID

Example #2 Calculations Involving pH and pOH Calculate [H 3 O + ], pH, [OH - ], and pOH for a M HNO 3 solution.

Example #2 Solution HNO 3 (aq) + H 2 O  H 3 O + (aq) + NO 3 - (aq) M M

Example #2 Solution HNO 3 (aq) + H 2 O  H 3 O + (aq) + NO 3 - (aq) M 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 + ] = - log = - (-1.82) = 1.82

Example #2 Solution pH + pOH = 14.00

Example #2 Solution pH + pOH = pOH = pH

Example #2 Solution pH + pOH = pOH = pH = = 12.18

Example #2 Solution pH + pOH = pOH = pH = = pOH = - log [OH - ]

Example #2 Solution pH + pOH = pOH = pH = = pOH = - log [OH - ] = - log [OH - ]

Example #2 Solution pH + pOH = pOH = pH = = pOH = - log [OH - ] = - log [OH - ] [OH - ] = [2 nd ]log

Example #2 Solution pH + pOH = pOH = pH = = pOH = - log [OH - ] = - log [OH - ] [OH - ] = [ 2 nd ] log =

Example #2 Solution pH + pOH = pOH = pH = = pOH = - log [OH - ] = - log [OH - ] [OH - ] = [2 nd ]log = = 6.7 x M

pH of STRONG BASE

Example #3 Calculations Involving pH and pOH Calculate [H 3 O + ], pH, [OH - ], and pOH for a M Ca(OH) 2 solution.

Example #3 Solution Ca(OH) 2 (aq)  Ca 2+ (aq) + 2 OH - (aq)

Example #3 Solution Ca(OH) 2 (aq)  Ca 2+ (aq) + 2 OH - (aq) M 2 x M = M

Example #3 Solution Ca(OH) 2 (aq)  Ca 2+ (aq) + 2 OH - (aq) M 2 x M = 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 - ] = - log = - (-1.52) = 1.52

Example #3 Solution pH + pOH = 14.00

Example #3 Solution pH + pOH = pH = pOH

Example #3 Solution pH + pOH = pH = pOH = = 12.48

Example #3 Solution pH + pOH = pH = pOH = = pH = - log [H 3 O + ]

Example #3 Solution pH + pOH = pH = pOH = = pH = - log [H 3 O + ] = - log [[H 3 O + ]

Example #3 Solution pH + pOH = pH = pOH = = pH = - log [H 3 O + ] = - log [[H 3 O + ] [H 3 O + ] = [2 nd ]log

Example #3 Solution pH + pOH = pH = pOH = = pH = - log [H 3 O + ] = - log [[H 3 O + ] [H 3 O + ] = [2 nd ]log =

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

pK a pK a = - log K a

pK a pK a = - log K a The lower the pK a, the stronger the acid

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

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