Where did you get 5E-5.
Is this how you set it up?
0.01 x 0.05 = 5E-4
If so, is it supposed to be 5E-4 instead of 5E-5?
What is the equilibrium constant for a weak acid HA that dissociates 1% at an initial concentration of 0.05 M according to the following balanced chemical equation?
HA <----> H+ + A-
Chemistry- DrBob222 - DrBob222, Saturday, March 12, 2016 at 11:03pm
........HA <----> H+ + A-
I.....0.05........0.....0
C....-5E-5........5E-5..5E-5
E......?..........5E-5..5E-5
Substitute the E line (after evaluation 0.05-5E-5) and solve for Ka.
To determine the equilibrium constant (Ka) for the dissociation of the weak acid HA, we need to use the information provided in the chemical equation and the initial concentration of HA.
The dissociation equation is given as:
HA <----> H+ + A-
The initial concentration of HA is 0.05 M, and we are told that it dissociates 1%.
So, 1% of 0.05 M is 0.0005 M or 5E-4 M. This is the concentration of both H+ and A- at equilibrium.
To solve for Ka, we need to set up an ICE table (Initial, Change, Equilibrium).
Let's do that:
........HA <----> H+ + A-
I.....0.05........0.....0
C....-5E-5........5E-5..5E-5
E......?..........5E-5..5E-5
In the ICE table, "I" represents the initial concentrations, "C" represents the changes in concentration, and "E" represents the equilibrium concentrations.
The change in concentration for HA is -5E-5 M because it decreases as it dissociates.
For H+ and A-, the change in concentration is +5E-5 M because they are formed as a result of the dissociation.
At equilibrium, the concentration of H+ and A- is 5E-5 M each.
The equilibrium constant (Ka) expression for this reaction is:
Ka = [H+][A-] / [HA]
Plugging in the equilibrium concentrations:
Ka = (5E-5)(5E-5) / (0.05 - 5E-5)
Now, we can solve for Ka using a calculator or by multiplying and dividing the numbers:
Ka = 2.5E-9.
So, the equilibrium constant (Ka) for the dissociation of the weak acid HA is 2.5E-9.
To calculate the equilibrium constant (Ka) for the weak acid HA that dissociates 1% at an initial concentration of 0.05 M, we can use the following steps:
1. Write the balanced equation for the dissociation of HA:
HA <----> H+ + A-
2. Set up the initial, change, and equilibrium concentrations in a table:
HA H+ A-
I: 0.05 0 0
C: -5E-5 5E-5 5E-5
E: ? 5E-5 5E-5
Here, we assume that the initial concentration of HA is 0.05 M and that it dissociates by 1%.
Since 1% of 0.05 M is 0.0005 M (5E-5), we subtract 5E-5 from the initial concentration of HA in the change row.
3. Now, we substitute the E line (after evaluating 0.05 - 5E-5) to determine the value of the equilibrium concentration of HA. In this case, we solve for the equilibrium concentration of HA (denoted by "?").
0.05 - 5E-5 = ?
? = 0.04995
Therefore, the equilibrium concentration of HA is approximately 0.04995 M.
4. Now, we can calculate the equilibrium constant (Ka) using the equation:
Ka = [H+][A-] / [HA]
Given that the equilibrium concentrations of H+ and A- are both 5E-5 M and the equilibrium concentration of HA is 0.04995 M, we can substitute these values into the equation:
Ka = (5E-5)(5E-5) / (0.04995)
After solving this equation, you can obtain the value of the equilibrium constant (Ka) for the weak acid HA.