@DrBob222
I got the last question that I posted wrong and I solved it the same way. Could you show me the answer and how you got it? My test is Friday and this is a practice problem.
Suppose that 30 mL of 0.5 M CsOH and 55 mL of 0.1 M NaHS are mixed.
What is the K for the dominant equilibrium in terms of Ka's, Kb's, Kw etc.?
Ka of HS-
1/Ka of HS-
Kb of OH-
1/Kb of OH-
Kb of S-2
1/Kb of S-2
Kw
1/Kw
There is no reaction.
Post your work and let me check it.
To find the K for the dominant equilibrium in this mixture, we need to determine which species will undergo significant reaction and form the dominant equilibrium.
First, let's write the balanced chemical equations for the dissociation of the given compounds:
CsOH -> Cs+ + OH-
NaHS -> Na+ + HS-
Now, we need to consider which species will undergo significant reaction and contribute to the dominant equilibrium. In this case, we can see that OH- and HS- will react to form water (H2O) and S2-. This is because OH- is a strong base and HS- is a weak acid.
So, the balanced chemical equation for the dominant equilibrium is:
OH- + HS- -> H2O + S2-
Now, let's look at the equilibrium constant expression for this equilibrium:
K = ([H2O][S2-]) / ([OH-][HS-])
Since water (H2O) is in its liquid state and pure liquids and solids are not included in the equilibrium expression, we can simplify the expression as follows:
K = ([S2-]) / ([OH-][HS-])
Now, let's substitute the concentrations of the ions into the equation:
[OH-] = 0.5 M (from 30 mL of 0.5 M CsOH)
[HS-] = 0.1 M (from 55 mL of 0.1 M NaHS)
K = ([S2-]) / (0.5 M * 0.1 M)
= ([S2-]) / (0.05 M)
Therefore, the K for the dominant equilibrium in terms of Ka's, Kb's, Kw, etc., is:
K = ([S2-]) / (0.05 M)
Since there is no reaction, the concentration of S2- is zero. Thus, the K value for the dominant equilibrium is also zero.
I hope this explanation helps you understand how to approach and solve this type of problem. Good luck with your test!