To determine the pOH of a 0.24 M solution of potassium nitrite (KNO2), we need to understand how salts like KNO2 behave in water.
Potassium nitrite is a salt derived from a strong base (KOH) and a weak acid (HNO2). When KNO2 dissolves in water, it dissociates into potassium ions (K⁺) and nitrite ions (NO2⁻). The potassium ion does not hydrolyze significantly because it is the cation from a strong base. However, the nitrite ion is the conjugate base of the weak acid nitrous acid (HNO2) and will hydrolyze to a certain extent.
Step 1: Identify the hydrolysis reaction
The nitrite ion (NO2⁻) can hydrolyze according to the following reaction:
\[ \text{NO}_2^- + \text{H}_2\text{O} \rightleftharpoons \text{HNO}_2 + \text{OH}^- \]
This reaction produces OH⁻ ions, resulting in a basic solution.
Step 2: Write the equilibrium expression
Using the equilibrium constant expression for the hydrolysis of the nitrite ion, we can use the Kb value. To find Kb, we can use the relationship between Ka (the acid dissociation constant of HNO2) and Kb:
\[ K_w = K_a \cdot K_b \]
For nitrous acid (HNO2), the Ka value is approximately \( 4.5 \times 10^{-4} \) at room temperature. Therefore:
\[ K_b = \frac{K_w}{K_a} = \frac{1.0 \times 10^{-14}}{4.5 \times 10^{-4}} \approx 2.22 \times 10^{-11} \]
Step 3: Set up ICE table
Let's consider the hydrolysis equilibrium:
- Initial concentration of NO2⁻ = 0.24 M (from KNO2)
- Change = -x (for NO2⁻) and +x (for OH⁻)
- Equilibrium concentration of NO2⁻ = 0.24 - x, and OH⁻ = x
At equilibrium:
\[ K_b = \frac{[\text{HNO}_2][\text{OH}^-]}{[\text{NO}_2^-]} = \frac{x^2}{0.24 - x} \]
Since Kb is small, we can assume that \( x \) is much smaller than 0.24, thus \( 0.24 - x \approx 0.24 \).
Thus, the equation becomes:
\[ K_b = \frac{x^2}{0.24} \]
Step 4: Solve for x
Plug in the value of Kb:
\[ 2.22 \times 10^{-11} = \frac{x^2}{0.24} \]
Rearranging gives:
\[ x^2 = 2.22 \times 10^{-11} \times 0.24 \] \[ x^2 = 5.328 \times 10^{-12} \] \[ x = \sqrt{5.328 \times 10^{-12}} \approx 2.31 \times 10^{-6} , \text{M} \]
This \( x \) represents the concentration of hydroxide ions (\( [\text{OH}^-] \)).
Step 5: Calculate pOH
Calculate the pOH:
\[ \text{pOH} = -\log [\text{OH}^-] \] \[ \text{pOH} = -\log (2.31 \times 10^{-6}) \]
Calculating the logarithm:
\[ \text{pOH} \approx 5.64 \]
Conclusion
The pOH of a 0.24 M solution of potassium nitrite (KNO2) is approximately 5.64.