Good day,
In relation to equilibrium, how would you know if an acid would spontaneously dissociate?
For example in this problem:
The equilibrium constant for the reaction HNO2(aq) + H2O(ℓ) → NO− 2 (aq) + H3O+(aq) is 4.3 × 10−4 at 25◦ C. Will nitrous acid spontaneously dissociate when [HNO2(aq)] = 0.15 M and [NO− 2 (aq)] = [H3O+(aq)] = 1.0 × 10−2 M?
Of course it will. The equilibrium constant tells you it will.
Keq = (products)/reactants = 4.3E-4. In order for there to be no dissociation it would be necessary for K to be zero and it isn't.
Ohh! Thank you, sir!
To determine if an acid will spontaneously dissociate, you can compare the equilibrium constant (K) with the concentrations of the reactants and products at a given temperature.
In this problem, the equilibrium constant (K) for the reaction HNO2(aq) + H2O(ℓ) → NO−2(aq) + H3O+(aq) is given as 4.3 × 10−4 at 25°C.
To determine if nitrous acid (HNO2) will spontaneously dissociate, you need to compare the concentrations of the reactants ([HNO2] and [H2O]) and the products ([NO−2] and [H3O+]) with the given equilibrium constant (K).
The concentration of HNO2 is given as [HNO2(aq)] = 0.15 M, and the concentrations of [NO−2(aq)] and [H3O+(aq)] are given as 1.0 × 10−2 M for both.
Now, you can set up the equilibrium expression using the given concentrations:
K = [NO−2(aq)][H3O+(aq)] / [HNO2(aq)][H2O(ℓ)]
Substituting the given values:
4.3 × 10−4 = (1.0 × 10−2)(1.0 × 10−2) / (0.15)([H2O(ℓ)])
To find the concentration of water ([H2O(ℓ)]), you can rearrange the equation:
[H2O(ℓ)] = (1.0 × 10−2)(1.0 × 10−2) / (0.15)(4.3 × 10−4)
Simplifying the equation:
[H2O(ℓ)] = 0.17 M
Since the concentration of water is not negligible (0.17 M), it means that the reaction is not at the ideal conditions for spontaneous dissociation of nitrous acid. The concentration of water affects the equilibrium position, and in this case, the acid will not fully dissociate.
Good day!
To determine if an acid will spontaneously dissociate, we need to compare the reaction quotient (Q) to the equilibrium constant (K). If Q is less than K, the reaction is not at equilibrium and the acid will spontaneously dissociate. However, if Q is equal to or greater than K, the reaction is at equilibrium and the acid will not spontaneously dissociate.
To calculate Q, we need to write the expression for the reaction quotient using the concentrations of the species involved. In this case, the reaction is:
HNO2(aq) + H2O(l) → NO−2(aq) + H3O+(aq)
The expression for Q is as follows:
Q = ([NO−2] *[H3O+]) / ([HNO2] *[H2O])
Now we can plug in the given concentrations into the equation:
Q = ((1.0 × 10−2 M) * (1.0 × 10−2 M)) / ((0.15 M) * (1 M))
Q = 0.00667
To determine if nitrous acid will spontaneously dissociate, we compare Q to the given equilibrium constant (K = 4.3 × 10−4). Since Q (0.00667) is greater than K (4.3 × 10−4), it means that the reaction is at equilibrium or has shifted to the left.
Therefore, in this particular case, nitrous acid will not spontaneously dissociate when [HNO2(aq)] = 0.15 M and [NO−2(aq)] = [H3O+(aq)] = 1.0 × 10−2 M.
I hope this explanation helps! Let me know if there's anything else I can assist you with.