How do i fill out an ICE chart when i am only given an initial concentration of one reactants and an equilibrium of another reaction?
inital 0.045 HNO2(aq)
Equilibrium 0.0043 H3O+(aq)
NO2(aq) - no information
my first thought was to sqaure root .0043 mol/l by 10 to find the change in H3O.
also should intial of H3O and NO2 be zero?
How do i find pH of HNO2 i just need a formula to follow cause i cant figure out the equation.
..........HNO2 ==> H^+ + NO2^-
I.........0.045M....0.....0
C.........-x........x.....x
E........0.045-x....x.....x
You know x = (H^+) = 0.0043M because the problem tells you this is the equilibrium concn of H3O^+; therefore,
pH = -log(H^+) = ?
BTW, the problem doesn't tell you the concn of NO2^-; however, you know it is the same as H^+. You also know (HNO2) at equilibrium; it is 0.045-x = ? Finally, the problem doesn't ask for Ka but you can calculate it from
Ka = (H^+)(NO2^-)/(HNO2)
isnt the equilibrium of HNO2 0.0407 because the equilibrium of H+ is 0.0043?
i am confused if no change occured in the reaction according to the chart...
To fill out an ICE chart when you are given the initial concentration of one reactant and the equilibrium concentration of another, follow these steps:
1. Write down the balanced chemical equation for the reaction. From your question, it seems like the reaction is:
HNO2(aq) + H2O(l) ⇌ H3O+(aq) + NO2(aq)
2. Identify the initial concentration, equilibrium concentration, and change in concentration for the given reactants and products. In this case, you are given the initial concentration of HNO2(aq) and the equilibrium concentration of H3O+(aq). You are not given any information about the concentration of NO2(aq).
3. Set up the ICE chart as follows:
Species | Initial Concentration | Change in Concentration | Equilibrium Concentration
-------------------------------------------------------------------------------------------
HNO2(aq) | 0.045 | ? | ?
H3O+(aq) | ? | 0.0043 | ?
NO2(aq) | ? | ? | ?
4. Since you are not given the concentration of NO2(aq), leave it as a question mark in both the initial concentration and equilibrium concentration columns.
5. To find the change in concentration for H3O+(aq), you cannot simply take the square root of 0.0043 mol/L. The change in concentration is determined by the stoichiometry of the reaction. Since the stoichiometric coefficient for H3O+(aq) is 1 (meaning it appears as 1 H3O+ ion per molecule of HNO2), the change in concentration for H3O+(aq) will also be the same as the change in concentration for HNO2(aq).
6. To calculate the change in concentration for HNO2(aq) and H3O+(aq), you need to use the equilibrium constant expression and the information about initial concentrations.
7. The equilibrium constant expression for the reaction is:
K = [H3O+(aq)] * [NO2(aq)] / [HNO2(aq)]
8. Since the initial concentration of HNO2(aq) is given as 0.045 M, you can consider it as the equilibrium concentration since we don't have any information about the change in concentration.
9. Substitute the given equilibrium concentrations and the unknown change in concentration into the equilibrium constant expression.
0.0043 = (change in concentration) * [NO2(aq)] / 0.045
10. Rearrange the equation to solve for the change in concentration:
(change in concentration) = 0.0043 * 0.045 / [NO2(aq)]
11. Since we don't have information about [NO2(aq)], you cannot determine the exact value of the change in concentration for HNO2(aq) and H3O+(aq). But you can calculate their relative values if you assume a value for [NO2(aq)] or use the ICE chart to calculate equilibrium concentrations.
In summary, to fill out the ICE chart, set up the initial concentration for HNO2(aq) as 0.045 M, the equilibrium concentration for H3O+(aq) as 0.0043 M, leave the initial and equilibrium concentrations of NO2(aq) as question marks, and calculate the change in concentration for HNO2(aq) and H3O+(aq) using the equilibrium constant expression.