Chemistry 1: Calculate pH of 0.150 M Acetic Acid (Ka = 1.8 x 10⁻⁵) with ICE Table

Exact Chemistry Midterm Problem

Calculate the $\text{pH}$ and percent ionization of a $0.150\text{ M}$ aqueous solution of acetic acid ($\text{CH}_3\text{COOH}$, $K_a = 1.8 \times 10^{-5}$).

Step-by-Step ICE Table Solution

1. Set up Equilibrium Expression:

$$ K_a = \frac{[\text{CH}_3\text{COO}^-][\text{H}_3\text{O}^+]}{[\text{CH}_3\text{COOH}]} = \frac{x^2}{0.150 - x} = 1.8 \times 10^{-5} $$

1. Apply 5% Approximation ($0.150 - x \approx 0.150$):

$$ x^2 = (1.8 \times 10^{-5})(0.150) = 2.70 \times 10^{-6} \implies x = [\text{H}_3\text{O}^+] = 1.643 \times 10^{-3}\text{ M} $$

1. Check Percent Ionization:

$$ \% = \frac{1.643 \times 10^{-3}}{0.150} \times 100\% = 1.10\% < 5\% \quad (\text{Valid}) $$

1. Compute $\text{pH}$:

$$ \text{pH} = -\log_{10}(1.643 \times 10^{-3}) = \mathbf{2.78} $$