Equilibrium Concentrations

Determination of KaK_a from Equilibrium Concentrations

  • Acetic acid example:

    • Equilibrium concentrations: [CH<em>3CO</em>2H]=0.0787M[CH<em>3CO</em>2H] = 0.0787 M and [H<em>3O+]=[CH</em>3CO2−]=0.00118M[H<em>3O^+] = [CH</em>3CO_2^-] = 0.00118 M.

Determination of KaK_a from pH

  • Nitrous acid example:

    • pH of 0.0516 M HNO2HNO_2 solution is 2.34.

    • [H3O+]=10−2.34=0.0046M[H_3O^+] = 10^{-2.34} = 0.0046 M

    • Equilibrium concentrations: [HNO<em>2]=0.0470M[HNO<em>2] = 0.0470 M, [H</em>3O+]=0.0046M[H</em>3O^+] = 0.0046 M, [NO2−]=0.0046M[NO_2^-] = 0.0046 M

    • K<em>a=[H</em>3O+][NO<em>2−][HNO</em>2]=(0.0046)(0.0046)(0.0470)=4.6×10−4K<em>a = \frac{[H</em>3O^+][NO<em>2^-]}{[HNO</em>2]} = \frac{(0.0046)(0.0046)}{(0.0470)} = 4.6 \times 10^{-4}

Calculating Equilibrium Concentrations in a Weak Acid Solution

  • Formic acid example:

    • Concentration of 0.534 M formic acid, Ka=1.8×10−4K_a = 1.8 \times 10^{-4}

    • Ka=x20.534−x=1.8×10−4K_a = \frac{x^2}{0.534-x} = 1.8 \times 10^{-4}

    • Assuming x is small, x=0.534×(1.8×10−4)=9.6×10−5=9.8×10−3Mx = \sqrt{0.534 \times (1.8 \times 10^{-4})} = \sqrt{9.6 \times 10^{-5}} = 9.8 \times 10^{-3} M

    • [H3O+]=0.0098M[H_3O^+] = 0.0098 M

    • pH=−log[H3O+]=−log(0.0098)=2.01pH = -log[H_3O^+] = -log(0.0098) = 2.01

Calculating Equilibrium Concentrations without Simplifying Assumptions

  • Sodium bisulfate example:

    • pH of a 0.50 M solution of HSO<em>4−HSO<em>4^-, K</em>a=1.2×10−2K</em>a = 1.2 \times 10^{-2}