Find the pH of a 0.010 M HNO2 solution.

Steps to Solve

1. Identify the acid dissociation constant (Ka) We need to look up the Ka value for nitrous acid (HNO2). The typical value is $K_a = 4.5 \times 10^{-4}$.

2. Write the dissociation equation The equation for the dissociation of nitrous acid in water is: $$ HNO_2 \leftrightarrow H^+ + NO_2^- $$

3. Set up the expression for Ka The expression for the acid dissociation constant is given by: $$ K_a = \frac{[H^+][NO_2^-]}{[HNO_2]} $$

Since this is a weak acid, we can let $x$ be the concentration of hydrogen ions produced. The initial concentration of HNO2 is 0.010 M. Therefore, at equilibrium:

• $[H^+] = x$
• $[NO_2^-] = x$
• $[HNO_2] = 0.010 - x$

4. Substituting for Ka Substituting into the expression for Ka yields: $$ 4.5 \times 10^{-4} = \frac{x \cdot x}{0.010 - x} \approx \frac{x^2}{0.010} $$ (Note: We can approximate $0.010 - x \approx 0.010$ since $x$ will be very small.)

5. Solve for x Rearranging gives us: $$ x^2 = 4.5 \times 10^{-4} \times 0.010 $$ Now, we calculate: $$ x^2 = 4.5 \times 10^{-6} $$

Taking the square root: $$ x = \sqrt{4.5 \times 10^{-6}} \approx 2.12 \times 10^{-3} $$

6. Calculate the pH Now, we can find the pH using the formula: $$ pH = -\log[H^+] $$ Substituting in our value for $x$: $$ pH = -\log(2.12 \times 10^{-3}) $$

Calculating the pH gives us: $$ pH \approx 2.67 $$