What is the [OH−] in the final solution prepared by mixing 20.0 mL of 0.050 M HCl with 30.0 mL of 0.10 MBa(OH)2?

To find the [OH-] in the final solution, we need to use the concept of stoichiometry and the balanced chemical equation for the reaction between HCl and Ba(OH)2.

The balanced chemical equation for the reaction is:

2HCl + Ba(OH)2 -> 2H2O + BaCl2

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

First, let's calculate the number of moles of HCl and Ba(OH)2 used in the reaction.

Moles of HCl = volume (in L) x concentration (in mol/L) = 0.020 L x 0.050 mol/L = 0.001 mol

Moles of Ba(OH)2 = volume (in L) x concentration (in mol/L) = 0.030 L x 0.10 mol/L = 0.003 mol

Since the stoichiometric ratio between HCl and Ba(OH)2 is 2:1, we can see that 0.001 mol of HCl reacts with 0.001 mol of Ba(OH)2.

Therefore, the remaining moles of Ba(OH)2 in the solution after the reaction is 0.003 mol - 0.001 mol = 0.002 mol.

Now, let's calculate the volume of the final solution.

Volume of final solution = volume of HCl + volume of Ba(OH)2 = 20.0 mL + 30.0 mL = 50.0 mL = 0.050 L

To find the concentration of [OH-] in the final solution, we divide the moles of Ba(OH)2 by the volume of the final solution.

[OH-] = moles of Ba(OH)2 / volume of final solution = 0.002 mol / 0.050 L = 0.040 mol/L

Therefore, the [OH-] in the final solution prepared by mixing 20.0 mL of 0.050 M HCl with 30.0 mL of 0.10 M Ba(OH)2 is 0.040 mol/L.