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Wize High School Grade 11 Chemistry Textbook > Acids and Bases

pH and pOH Calculations

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pH and pOH Calculations

  • In pure water, at 25°C the following chemical reaction takes place:
2H2O⇌H3O++OH−2H_2 O⇌H_3 O^++OH^-
  • We can classify aqueous solutions based on the concentration of the hydronium ions:
  • neutral solutions: [H3O+(aq)]=1×10−7mol/L[H_3O^+(aq)]=1\times10^{-7}mol/L
  • acidic solutions: [H3O+(aq)]>1×10−7mol/L[H_3O^+(aq)]>1\times10^{-7}mol/L
  • basic solutions: [H3O+(aq)]<1×10−7mol/L[H_3O^+(aq)]<1\times10^{-7}mol/L
  • From here we can define pH, or the "power of hydrogen":
pH=−log⁡[H3O+(aq)]pH=-\log\left[H_3O^+\left(aq\right)\right] and [H3O+(aq)]=10−pH\left[H_3O^+\left(aq\right)\right]=10^{-pH}
  • We can also define pOH:
pOH=−log⁡[OH−(aq)]pOH=-\log\left[OH^-\left(aq\right)\right] and [OH−(aq)]=10−pOH[OH^-(aq)]=10^{-pOH}
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Example: Calculating pH

If 0.2mol of HCl are dissolved in 1.8L of water, what is the pH?
Since HCl\text{HCl} is a strong acid, we will be adding 0.2 mols of H+\text{H}^+ to 1.8 L of water.

[H+]=0.2mol1.8L=0.11mol/L[\text{H}^+]=\dfrac{0.2mol}{1.8L}=0.11mol/L

pH=−log⁡[H+]=0.95pH=-\log[\text{H}^+]=0.95

Practice: Calculating pH

Calculate the pH of a 2.75 x 10-3 mol/L aqueous HNO3 solution. Give your answer to two decimal places.

Practice: Understanding pH

A solution with a pH of 1 has ____ the hydronium ion concentration compared to a solution with a pH of 4.

Practice: Solutions and pH

Calculate the pH of a solution prepared by adding 25.00mL of 0.100mol/L HCl with 75.00mL of 2.50x10-2mol/L HBr. Give your answer rounded to two decimal points.