15 ml of 0.5 M of HCl is added to a 100 ml sample of 0.2 M HNO2 (Ka for HNO2= 4.0 x 10^(-4)). What is the equilibrium concentration of NO2- ions?
To find the equilibrium concentration of NO2- ions, we need to analyze the reaction between HCl and HNO2 and use the concept of acid-base equilibrium.
The balanced chemical equation for the reaction between HCl and HNO2 is:
HCl + HNO2 ⇌ NO2- + H2O
Let's denote the initial concentration of HNO2 as [HNO2]₀ and the change in concentration of HNO2 as Δ[HNO2]. The concentration of HCl is given as 0.5 M and the volume is 15 ml, which can be converted to 0.015 L by dividing by 1000.
First, we need to determine the concentration of HNO2 after the reaction occurs. Since we know the initial concentration of HNO2 is 0.2 M and the volume is 100 ml, we can convert it to 0.1 L by dividing by 1000.
[HNO2]₀ = 0.2 M
Volume of HNO2 solution = 100 ml = 0.1 L
Next, let's determine the moles of HNO2 initially present:
moles of HNO2 = [HNO2]₀ × volume
= 0.2 M × 0.1 L
= 0.02 mol
Now, let's determine the moles of HNO2 that react with HCl. Since the stoichiometric coefficient of HNO2 in the balanced equation is 1, the moles of HNO2 reactant that will be used up will be equal to the moles of HCl added, which can be calculated using the concentration and volume of HCl:
moles of HCl = concentration × volume
= 0.5 M × 0.015 L
= 0.0075 mol
Since the reaction between HCl and HNO2 is a 1:1 ratio, the moles of HNO2 that react with HCl is also 0.0075 mol.
Now, let's determine the final concentration of HNO2 after the reaction. Since the volume of the solution remains the same (0.1 L), we can use the formula:
[HNO2] = (moles of HNO2 remaining) / volume
= (moles of HNO2 initially) - (moles of HNO2 reacted) / volume
= (0.02 mol - 0.0075 mol) / 0.1 L
= 0.0125 mol / 0.1 L
= 0.125 M
Finally, using the equilibrium expression for the ionization of HNO2:
Ka = [NO2-][H3O+] / [HNO2]
Since we know the value of Ka (4.0 x 10^(-4)) and the concentration of HNO2 at equilibrium (0.125 M), we can rearrange the equation to find the concentration of NO2-:
[NO2-] = (Ka × [HNO2]) / [H3O+]
= (4.0 x 10^(-4)) × (0.125 M) / [H3O+]
Unfortunately, we are not given the concentration of H3O+ (H+). If we assume that the reaction between HCl and HNO2 goes to completion, HCl is a strong acid and dissociates fully to produce H3O+ ions. Thus, the concentration of H3O+ ions will be equal to the concentration of HCl (0.5 M).
Plugging in the given values, we can calculate the equilibrium concentration of NO2-:
[NO2-] = (4.0 x 10^(-4)) × (0.125 M) / (0.5 M)
= 8.0 x 10^(-5) M
Therefore, the equilibrium concentration of NO2- ions is 8.0 x 10^(-5) M.
To find the equilibrium concentration of NO2- ions, we need to set up an equation based on the reaction between HNO2 and HCl.
The balanced chemical equation for the reaction is:
HNO2 + HCl ⇌ NO2- + H3O+
From the equation, we can see that 1 mole of HNO2 reacts with 1 mole of HCl to form 1 mole of NO2- ions.
Given:
Volume of HCl (V1) = 15 ml = 0.015 L
Molarity of HCl (M1) = 0.5 M
Volume of HNO2 (V2) = 100 ml = 0.1 L
Molarity of HNO2 (M2) = 0.2 M
Ka for HNO2 = 4.0 x 10^(-4)
To calculate the equilibrium concentration of NO2- ions, we need to consider the reaction stoichiometry and the initial concentrations of both HNO2 and HCl.
1. Calculate the moles of HNO2:
Moles of HNO2 = Molarity of HNO2 × Volume of HNO2
= 0.2 M × 0.1 L
= 0.02 moles
2. Calculate the moles of HCl:
Moles of HCl = Molarity of HCl × Volume of HCl
= 0.5 M × 0.015 L
= 0.0075 moles
3. Based on the reaction stoichiometry, we can determine that the moles of NO2- ions formed will be equal to the moles of HNO2 reacted (since the ratio is 1:1).
Moles of NO2- = Moles of HNO2 reacted = Moles of HCl
= 0.0075 moles
4. Calculate the equilibrium concentration of NO2- ions by dividing the moles of NO2- by the total volume:
Equilibrium concentration of NO2- = Moles of NO2- / Total volume
= 0.0075 moles / (0.1 L + 0.015 L)
= 0.0075 moles / 0.115 L
≈ 0.0652 M
Therefore, the equilibrium concentration of NO2- ions is approximately 0.0652 M.