Suppose that 50 mL of 0.3 M LiOH and 60 mL of 0.1 M NaHSO3 are mixed.

What is the K for the dominant equilibrium in terms of Ka's, Kb's, Kw etc.?

Ka of HSO3-
1/Ka of HSO3-
Kb of OH-
1/Kb of OH-
Kb of SO3-2
1/Kb of SO3-2
Kw
1/Kw
There is no reaction.

LiOH + NaHSO3 ==> H2O + NaLiSO3

You have 50*0.3 = 15 mmoles LiOH
You have 50*0.1 = 5 mmoles NaHSO3
So you have 10 mmoles LiOH in excess (which is a strong base) with all of the NaHSO3 reacted. Thus the dominant equilibrium, as I see it, is the strong base of LiOH.

To determine the K for the dominant equilibrium, we first need to identify the possible equilibrium reactions that can occur when LiOH and NaHSO3 are mixed.

The reaction between LiOH and NaHSO3 can proceed in two ways:

1) LiOH can react with HSO3- to form LiHSO3 and H2O.
2 LiOH + HSO3- --> LiHSO3 + H2O

2) LiOH can react with HSO3- to form LiSO3 and H2O.
LiOH + HSO3- --> LiSO3 + H2O

Let's calculate the concentration of LiHSO3, LiSO3, and OH- after the reaction.

Step 1: Calculate the moles of LiOH and NaHSO3:

Moles of LiOH = Volume of LiOH solution (50 mL) × Concentration of LiOH solution (0.3 M)
Moles of LiOH = 0.05 L × 0.3 mol/L = 0.015 mol

Moles of NaHSO3 = Volume of NaHSO3 solution (60 mL) × Concentration of NaHSO3 solution (0.1 M)
Moles of NaHSO3 = 0.06 L × 0.1 mol/L = 0.006 mol

Step 2: Determine the limiting reactant:

The limiting reactant is the one that is completely consumed first. We need to compare the moles of LiOH and NaHSO3 to determine the limiting reactant.

The stoichiometric ratio between LiOH and HSO3- is 1:1 in both reactions. Since the moles of NaHSO3 (0.006 mol) are smaller than the moles of LiOH (0.015 mol), NaHSO3 is the limiting reactant.

Step 3: Calculate the moles and concentrations of the products:

From the reaction equation, we can see that 1 mole of NaHSO3 produces 1 mole of either LiHSO3 or LiSO3.

Moles of LiHSO3 formed = Moles of NaHSO3 (0.006 mol)
Concentration of LiHSO3 = Moles of LiHSO3 / Total volume of the solution (50 mL + 60 mL)

Moles of LiSO3 formed = Moles of NaHSO3 (0.006 mol)
Concentration of LiSO3 = Moles of LiSO3 / Total volume of the solution (50 mL + 60 mL)

Step 4: Determine the concentration of OH-:

Since LiOH is a strong base, it completely dissociates into Li+ and OH-. The initial concentration of LiOH is 0.3 M, so the concentration of OH- is also 0.3 M.

Now we can write the expression for K of the dominant equilibrium:

K = [LiHSO3] / [OH-]

Where [LiHSO3] is the concentration of LiHSO3 and [OH-] is the concentration of OH-, both determined in Step 3 and Step 4.

So, in terms of Ka's, Kb's, and Kw:

K = [LiHSO3] / [OH-]
Ka of HSO3- = 1 / Kb of HS03-
Kb of OH- = Kw / Ka of OH-
Kb of SO3-2 = 1 / Ka of SO3-2
Kw = [H+][OH-]
There is no reaction, K = 0

Please note that the values of Ka and Kb need to be provided to calculate the equilibrium constants accurately.