The problem states that we need to determine the concentration of OH- in the final solution prepared by mixing 20 ml of 0.05 molar HCl and 30 ml of 0.1 molar Ba(OH)2. To solve this problem, we will follow these steps:

1. Write down the balanced chemical equation for the reaction between HCl and Ba(OH)2. The balanced equation is as follows:
HCl + Ba(OH)2 → BaCl2 + 2H2O

2. Identify the stoichiometry of the reaction. From the balanced equation, we can see that 1 mole of HCl reacts with 1 mole of Ba(OH)2 to produce 1 mole of BaCl2 and 2 moles of water.

3. Calculate the number of moles of HCl and Ba(OH)2 used in the reaction.
Moles of HCl = volume (in L) × molarity = 0.02 L × 0.05 mol/L = 0.001 mol
Moles of Ba(OH)2 = volume (in L) × molarity = 0.03 L × 0.1 mol/L = 0.003 mol

4. Determine the limiting reagent. The limiting reagent is the reactant that is completely consumed in the reaction and determines the maximum amount of product that can be formed. In this case, HCl and Ba(OH)2 have a 1:1 stoichiometry, so the limiting reagent is the one that is present in the smallest amount. Since HCl has fewer moles (0.001 mol) compared to Ba(OH)2 (0.003 mol), HCl is the limiting reagent.

5. Calculate the moles of OH- ions produced from the reaction. Since HCl is the limiting reagent, it will react completely with Ba(OH)2. From the balanced equation, we can see that 1 mole of Ba(OH)2 produces 2 moles of OH- ions. Therefore, the moles of OH- ions produced will be twice the moles of Ba(OH)2 used in the reaction.
Moles of OH- ions = 2 × moles of Ba(OH)2 = 2 × 0.003 mol = 0.006 mol

6. Calculate the concentration of OH- ions in the final solution. We need to determine the final volume of the solution by adding the volumes of HCl and Ba(OH)2 used.
Final volume = 20 ml + 30 ml = 50 ml = 0.05 L

Concentration of OH- ions = moles of OH- ions / final volume
Concentration of OH- ions = 0.006 mol / 0.05 L = 0.12 mol/L

Therefore, the concentration of OH- ions in the final solution prepared by mixing 20 ml of 0.05 molar HCl and 30 ml of 0.1 molar Ba(OH)2 is 0.12 mol/L.