Acids and Bases - Arrhenius Definition and pH/pOH Calculations

Arrhenius Definition of Acids and Bases

  • Two theories for defining acids and bases; Arrhenius definition and another in grade 12.

  • Arrhenius theory states that some compounds form electrically charged particles (ions) when in solution.

Acids

  • Acids ionize, meaning they break into ions. Sulfuric acid (H2SO4)insolutionbreaksdownintohydrogenions() in solution breaks down into hydrogen ions (H^+)andsulfateions() and sulfate ions (SO4^{2-}.

  • Ionization is when a molecular compound breaks into ions.

  • Acids increase the concentration of hydrogen ions (H+H^+) in solution.

    • [H+][H^+] refers to the concentration of hydrogen ions.

Bases

  • Bases dissociate, meaning an ionic compound breaks down into ions.

  • Dissociation means to split into separate parts.

  • Bases increase the concentration of hydroxide ions (OH−OH^-) in solution.

Acids and Bases in Solution

  • Acids and bases break apart into ions when put in solution, enabling them to conduct electricity.

  • These ions (hydrogen and hydroxide) didn't exist prior to ionization or dissociation.

  • Differentiating between dissociation and ionization involves identifying the type of bonds or compound.

Hydrogen Ions and Acidity/Basicity

  • High concentration of hydrogen ions (H+H^+) indicates an acidic solution.

  • High concentration of hydroxide ions (OH−OH^-) indicates a basic solution.

  • Equal concentrations of hydrogen and hydroxide ions indicate a neutral solution.

Limitations of the Arrhenius Theory

  • Hydrogen ion (H+H^+) is essentially a proton and is unstable, readily reacting with water molecules to form hydronium ion (H3O+H_3O^+). This is further discussed in grade 12.

  • The theory doesn't explain why compounds without hydroxide ions, like ammonia (NH3NH_3), can still have basic properties.

pH Scale

  • The pH scale measures hydrogen ion concentration ([H+][H^+]).

  • Ranges from 0 to 14.

  • pH is calculated using the formula: pH=−log[H+]pH = -log[H^+].

  • All formulas are available on the periodic table.

  • Moving towards 0 increases hydrogen concentration by tenfold, indicating a stronger acid.

  • Moving towards 14 decreases hydrogen concentration, indicating a stronger base.

Significant Figures in pH

  • When reporting pH values, only the numbers to the right of the decimal place count as significant figures.

  • Example: If [H+]=1×10−10[H^+] = 1 \times 10^{-10}, pH=10.0pH = 10.0 (one significant digit after the decimal).

  • The number of decimal places in the pH value corresponds to the number of significant figures in the hydrogen ion concentration.

  • When converting from pH to hydrogen ion concentration, the number of decimal places in the pH value determines the number of significant digits in the concentration.

  • Given pH = 4.0, the [H+H^+] = 10−410^{-4} M. Because the pH has one decimal place, the concentration should be expressed with only one significant figure.

  • Formula for calculating hydrogen ion concentration from pH: [H+]=10−pH[H^+] = 10^{-pH}.

Example Calculations

  • If [H+]=1×10−10[H^+] = 1 \times 10^{-10}, pH=−log(1×10−10)=10.0pH = -log(1 \times 10^{-10}) = 10.0. (1 sig fig)

  • If [H+]=1.7×10−2[H^+] = 1.7 \times 10^{-2}, pH=−log(1.7×10−2)=1.77pH = -log(1.7 \times 10^{-2}) = 1.77. (2 sig figs)

  • If [H+]=2.22×10−7[H^+] = 2.22 \times 10^{-7}, pH=−log(2.22×10−7)=6.653pH = -log(2.22 \times 10^{-7}) = 6.653. (3 sig figs)

  • If [H+]=9.6×10−6[H^+] = 9.6 \times 10^{-6}, pH=−log(9.6×10−6)=5.02pH = -log(9.6 \times 10^{-6}) = 5.02. (2 sig figs)

pOH Scale

  • pOH measures hydroxide ion concentration ([OH−][OH^-]).

  • Formula: pOH=−log[OH−]pOH = -log[OH^-].

  • On the pOH scale, 14 indicates low hydroxide concentration, and 0 indicates high hydroxide concentration.

  • The relationship between pH and pOH: pH+pOH=14pH + pOH = 14.

Calculations with pOH

  • Calculating pOH from hydroxide ion concentration: pOH=−log[OH−]pOH = -log[OH^-].

  • If [OH−]=6.2×10−3[OH^-] = 6.2 \times 10^{-3}, pOH=−log(6.2×10−3)=2.208pOH = -log(6.2 \times 10^{-3}) = 2.208 (2 sig figs -> 2 decimal places).

  • Calculating hydroxide ion concentration from pOH: [OH−]=10−pOH[OH^-] = 10^{-pOH}.

Example Calculations

  • If [OH−]=7.9×10−2[OH^-] = 7.9 \times 10^{-2}, pOH=−log(7.9×10−2)=1.003pOH = -log(7.9 \times 10^{-2}) = 1.003 (2 sig figs given therefore round to 2 decimal places for pOH).

  • If [OH−]=6.22×10−2[OH^-] = 6.22 \times 10^{-2}, pOH=−log(6.22×10−2)=1.206pOH = -log(6.22 \times 10^{-2}) = 1.206 (3 sig figs --> 3 decimal places).

  • If [OH−]=9.411×10−6[OH^-] = 9.411 \times 10^{-6}, pOH=−log(9.411×10−6)=5.026pOH = -log(9.411 \times 10^{-6}) = 5.026 (4 sig figs --> 4 decimal places).

  • If [OH−]=2.0×10−9[OH^-] = 2.0 \times 10^{-9}, pOH=−log(2.0×10−9)=8.70pOH = -log(2.0 \times 10^{-9}) = 8.70 (2 sig figs --> two decimal places).

  • If pOH is 13.0, [OH−]=1×10−13M[OH^-] = 1 \times 10^{-13} M (1 decimal place in pOH, so 1 sig fig in the final concentration).

  • If pOH is 5.61, the [OH−]=2×10−6[OH^-] = 2 \times 10^{-6} M (2 sig figs).

  • If pOH is 2.714, the [OH−]=1.93×10−3M[OH^-] = 1.93 \times 10^{-3} M (3 sig figs).

  • If pOH is 8.2, then [OH−]=6×10−9M[OH^-] = 6 \times 10^{-9} M (1 sig fig).