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pH and pOH, Exercises of Chemistry

The University of Texas at AustinChemistry

pH = negative logarithm of the molar concentration of hydronium ion. • pOH = negative logarithm of molar concentration of hydroxide ion. pH = -log[H3O+].

Typology: Exercises

2022/2023

Uploaded on 02/28/2023

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eekanath 🇺🇸

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pH and pOH
•when working with dilute solutions of strong
acids or weak acids (likewise with bases)
> [H3O+] or [OH-] is very small, often around
10-6 or smaller
> with such small quantities, it is difficult to
compare concentrations
> pH and pOH scales developed
•pH = negative logarithm of the molar
concentration of hydronium ion
•pOH = negative logarithm of molar
concentration of hydroxide ion
pH = -log[H3O+]
pOH = -log[OH-]
pf3
pf4
pf5

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pH and pOH

  • when working with dilute solutions of strong acids or weak acids (likewise with bases) > [H 3 O + ] or [OH - ] is very small, often around 10
  • 6 or smaller > with such small quantities, it is difficult to compare concentrations > pH and pOH scales developed
  • pH = negative logarithm of the molar concentration of hydronium ion
  • pOH = negative logarithm of molar concentration of hydroxide ion pH = - log[H 3 O

] pOH = - log[OH

]

logarithm = the exponent to which 10 must be raised to represent a certain number For example... log 100 means 10 to the power of WHAT equals 100? log 100 = 2 because 10 2 = 100 log 1000 = log( 0. 01 ) = The reverse procedure of log is called antilog. Antilog is equivalent of raising 10 to the power of a certain exponent. antilog (x) = 10 x What is the antilog of 2? antilog ( 2 ) = 10 2 = 100

Consider the following: pH = - log[H 3 O

]

  • if we want to rearrange the equation to solve for [H 3 O + ], first remove the negative sign by multiplying both sides by - 1 to give: - pH = log[H 3 O + ]
  • since the reverse of log is antilog, the [H 3 O
    • ] is taking the antilog, the [H 3 O
      • ] can be determined by taking the antilog of both sides of the equation: antilog (-pH) = antilog (log[H 3 O
        • ]) = [H 3 O + ] antilog (-pH) = [H 3 O

] [H 3 O

] = 10

  • pH

  • similarly, [OH

     ] can be determined from the pOH

[H 3 O

] = antilog(-pH) = 10

  • pH [OH

] = antilog(-pOH) = 10

  • pOH Example: a) If the pH is 3. 17 , what is the [H 3 O

]? b) If the pOH is 5. 32 , what is the [OH

]?

There is a simple but important relationship between pH, pOH and Kw.

  • starting with Kw: [H 3 O + ][OH - ] = Kw
  • take the logarithm of both sides: log([[H 3 O + ][OH - ]) = logKw log[H 3 O

] + log[OH

] = logKw

  • multiply by - 1 : (-log[H 3 O + ]) + (-log[OH - ]) = (-logKw) pH + pOH = pKw
  • at 25 Kw = 1. 0 x 10
    • 14 pH + pOH = - log( 1. 0 X 10
  • 14 ) = 14. 00 pH + pOH = 14. 00

at 25 ℃: pH = 7. 0 is neutral pH < 7. 0 is acidic pH > 7. 0 is basic (alkaline)

  • since the pH and pOH scales are logarithmic scales, a difference in one pH or pOH unit is equivalent to a ten-fold ( 10 ×) difference in concentration > pH and pOH are the negative of the exponent, so low pH and pOH mean relatively high values of [H 3 O + ] and [OH - ] & high values of pH and pOH mean relatively low [H 3 O + ] and [OH - ]
  • negative pH values are possible for concentrated strong acids
  1. 00 M HCl has [H 3 O

] = 2. 00 mol/L pH = - log( 2. 00 ) = - 0. 30