To determine the pH of a 0.52 M nitrous acid (HNO2) solution with a given acid dissociation constant, [tex]\( K_a = 3.0 \times 10^{-8} \)[/tex], we can follow a detailed, step-by-step process.
### Step 1: Understand the Dissociation of HNO2
Nitrous acid (HNO2) is a weak acid and only partially dissociates in an aqueous solution.
[tex]\[ \text{HNO}_2 \rightleftharpoons \text{H}^+ + \text{NO}_2^- \][/tex]
### Step 2: Set Up the Equilibrium Expression
For the dissociation of a weak acid, the equilibrium constant expression (K_a) is given by:
[tex]\[ K_a = \frac{[\text{H}^+][\text{NO}_2^-]}{[\text{HNO}_2]} \][/tex]
Given:
- Concentration of nitrous acid, [tex]\([ \text{HNO}_2 ] = 0.52 \, \text{M}\)[/tex]
- Acid dissociation constant, [tex]\( K_a = 3.0 \times 10^{-8} \)[/tex]
### Step 3: Determine the Concentration of [tex]\( \text{H}^+ \)[/tex] Ions
Assume that the initial concentration of hydrogen ions ([tex]\( \text{H}^+ \)[/tex]) and nitrite ions ([tex]\( \text{NO}_2^- \)[/tex]) is negligible. As the acid dissociates, let the concentration of [tex]\( \text{H}^+ \)[/tex] and [tex]\( \text{NO}_2^- \)[/tex] each be [tex]\( x \)[/tex] at equilibrium. Therefore, the concentration of undissociated HNO2 is approximately [tex]\( 0.52 - x \)[/tex].
The equilibrium expression then becomes:
[tex]\[ K_a = \frac{x \cdot x}{0.52 - x} \][/tex]
Since [tex]\( K_a \)[/tex] is very small, we can assume [tex]\( x \ll 0.52 \)[/tex], so [tex]\( 0.52 - x \approx 0.52 \)[/tex].
Thus, the modified equilibrium expression is:
[tex]\[ 3.0 \times 10^{-8} = \frac{x^2}{0.52} \][/tex]
### Step 4: Solve for [tex]\( x \)[/tex]
Rearrange the above equation to solve for [tex]\( x \)[/tex]:
[tex]\[ x^2 = K_a \cdot 0.52 \][/tex]
[tex]\[ x^2 = 3.0 \times 10^{-8} \cdot 0.52 \][/tex]
[tex]\[ x^2 = 1.56 \times 10^{-8} \][/tex]
[tex]\[ x = \sqrt{1.56 \times 10^{-8}} \][/tex]
[tex]\[ x \approx 0.0001249 \][/tex]
Therefore, the concentration of [tex]\( \text{H}^+ \)[/tex], [tex]\( [ \text{H}^+ ] \approx 0.0001249 \, \text{M} \)[/tex].
### Step 5: Calculate the pH
The pH is calculated using the formula:
[tex]\[ \text{pH} = -\log[ \text{H}^+ ] \][/tex]
[tex]\[ \text{pH} = -\log(0.0001249) \][/tex]
[tex]\[ \text{pH} \approx 3.90 \][/tex]
### Conclusion
The pH of a 0.52 M nitrous acid solution is approximately 3.90.
