What is the percent dissociation of HN)2 when 0.064g of sodium nitrite is added to 120.0mL of a 0.057 M HNO2 solution? Ka for HNO2 is 4.0 x 10-4
The correct answer and explanation is :
To calculate the percent dissociation of HNO₂ (nitrous acid) when sodium nitrite (NaNO₂) is added to a solution, we need to carefully consider the equilibria involved. Here’s the step-by-step process for calculating the percent dissociation:
Step 1: Understanding the Problem
We are given:
- Mass of sodium nitrite (NaNO₂): 0.064 g
- Volume of HNO₂ solution: 120.0 mL = 0.120 L
- Concentration of HNO₂: 0.057 M
- Ka for HNO₂: 4.0 × 10⁻⁴
- Molar mass of NaNO₂: 69.0 g/mol (for NaNO₂)
Sodium nitrite dissociates in water to produce NO₂⁻ ions, which can influence the dissociation of HNO₂ because NO₂⁻ can act as a conjugate base and suppress the dissociation of HNO₂.
The reaction for HNO₂ dissociation is:
[
\text{HNO₂} \rightleftharpoons \text{H}^+ + \text{NO₂}^-
]
The nitrite ions from sodium nitrite will shift the equilibrium of HNO₂ dissociation, as the added NO₂⁻ ions will suppress HNO₂ dissociation according to Le Chatelier’s principle.
Step 2: Calculate Moles of NaNO₂
First, we calculate the moles of NaNO₂ added to the solution:
[
\text{Moles of NaNO₂} = \frac{\text{Mass of NaNO₂}}{\text{Molar Mass of NaNO₂}} = \frac{0.064 \, \text{g}}{69.0 \, \text{g/mol}} = 9.28 \times 10^{-4} \, \text{mol}
]
Since sodium nitrite dissociates completely in water, the moles of NO₂⁻ produced will be equal to the moles of NaNO₂ added, which is 9.28 × 10⁻⁴ mol.
Step 3: Concentration of NO₂⁻
Now, calculate the concentration of NO₂⁻ in the solution:
[
\text{Concentration of NO₂} = \frac{\text{Moles of NO₂⁻}}{\text{Volume of solution}} = \frac{9.28 \times 10^{-4} \, \text{mol}}{0.120 \, \text{L}} = 7.73 \times 10^{-3} \, \text{M}
]
Step 4: Set up the Equilibrium Expression
We now have a system in which HNO₂ dissociates, and NO₂⁻ is present. We can use the equilibrium expression for HNO₂ dissociation:
[
\text{Ka} = \frac{[\text{H}^+][\text{NO₂}^-]}{[\text{HNO₂}]}
]
Let ( x ) represent the change in concentration of HNO₂ that dissociates. At equilibrium, we have:
- ([HNO₂] = 0.057 – x)
- ([H^+] = x)
- ([NO₂^-] = 7.73 \times 10^{-3} + x) (since NO₂⁻ is also contributed by sodium nitrite)
Using the Ka expression, we get:
[
4.0 \times 10^{-4} = \frac{x(7.73 \times 10^{-3} + x)}{0.057 – x}
]
Step 5: Solve the Equation
Assume that ( x ) is small compared to 7.73 × 10⁻³ M and 0.057 M (which is a valid assumption due to the small value of Ka). This simplifies the equation to:
[
4.0 \times 10^{-4} = \frac{x(7.73 \times 10^{-3})}{0.057}
]
Solve for ( x ):
[
x = \frac{(4.0 \times 10^{-4})(0.057)}{7.73 \times 10^{-3}} = 2.95 \times 10^{-5} \, \text{M}
]
Step 6: Calculate the Percent Dissociation
The percent dissociation of HNO₂ is given by:
[
\text{Percent dissociation} = \frac{x}{\text{Initial concentration of HNO₂}} \times 100
]
[
\text{Percent dissociation} = \frac{2.95 \times 10^{-5}}{0.057} \times 100 = 0.052%
]
Final Answer:
The percent dissociation of HNO₂ is approximately 0.052%.
Explanation:
In this solution, the addition of sodium nitrite (which dissociates to form NO₂⁻) suppresses the dissociation of HNO₂. This is because NO₂⁻ is the conjugate base of HNO₂, and according to Le Chatelier’s principle, the presence of extra NO₂⁻ ions shifts the equilibrium to the left, reducing the extent of dissociation of HNO₂. This results in a very low percent dissociation, calculated to be about 0.052%.