Calculate the pH of 0.010M HNO2 solution. The Ka for HNO2 is 4.6 x 10^-4.
The Correct Answer and Explanation is :
To calculate the pH of a 0.010 M HNO2 solution, we need to consider the dissociation of nitrous acid (HNO2) in water. Since HNO2 is a weak acid, it partially dissociates into H+ (protons) and NO2- (nitrite ions). The dissociation equilibrium for HNO2 is:
[
HNO_2 \rightleftharpoons H^+ + NO_2^-
]
The acid dissociation constant, Ka, is given by:
[
K_a = \frac{[H^+][NO_2^-]}{[HNO_2]}
]
We are provided with:
- ( [HNO_2] = 0.010 \, \text{M} )
- ( K_a = 4.6 \times 10^{-4} )
Step 1: Set up the ICE table
We assume that the dissociation of HNO2 is small, so we can use an approximation for the concentrations at equilibrium.
| Species | Initial concentration (M) | Change in concentration (M) | Equilibrium concentration (M) |
|---|---|---|---|
| HNO2 | 0.010 | -x | 0.010 – x |
| H+ | 0 | +x | x |
| NO2- | 0 | +x | x |
Step 2: Write the expression for ( K_a )
[
K_a = \frac{x \cdot x}{0.010 – x} = 4.6 \times 10^{-4}
]
Since ( K_a ) is small, we assume ( 0.010 – x \approx 0.010 ), simplifying the equation:
[
K_a = \frac{x^2}{0.010}
]
Step 3: Solve for x
Now solve for ( x ), which represents the concentration of H+ ions:
[
4.6 \times 10^{-4} = \frac{x^2}{0.010}
]
[
x^2 = (4.6 \times 10^{-4})(0.010)
]
[
x^2 = 4.6 \times 10^{-6}
]
[
x = \sqrt{4.6 \times 10^{-6}} = 2.14 \times 10^{-3} \, \text{M}
]
Step 4: Calculate pH
The pH is given by:
[
\text{pH} = -\log[H^+]
]
Substitute the value of ( x ):
[
\text{pH} = -\log(2.14 \times 10^{-3}) = 2.72
]
Thus, the pH of the 0.010 M HNO2 solution is 2.72.
I will now generate an image of the calculation process.
Here is the diagram that shows the steps for calculating the pH of a 0.010 M HNO2 solution, including the ICE table, the acid dissociation constant equation, and the final pH calculation. It visually explains the entire process. Let me know if you need any further clarification!
