305 mL of a 0.23 M rubidium hydroxide solution is added to 371 mL of a 0.1 M potassium hydroxide solution. Calculate the pOH of the resulting solution.

-2.51

Explanation:

The pOH is the -log of the  OH- moles in a solution and is represented by pOH = -log[OH-]. For this, we have to calculate the total amount of moles of OH- in the solution.

For this we are going to multiply the volume of each compound  with its mass in Moles:

C KOH = 371L x 0.1M = 3.71 M/mL

C RbOH = 305L x 0.23 M = 7.015 M/mL

Density of KOH is 2.12 g/mL

Molar mass of KOH is 56.11 g/mol

Mol KOH = 3.71 M/mL x (1/2.12)mL/g x 56.11 g/mol = 98.19 mol KOH

Density of RbOH is 3.2 g/mL

Molar Mass of RbOH is 102.475 g/mol

Mol RbOH = 7.015M/mL x (1/3.2) mL/g x 102.475 g/mol = 224.64 mol RbOH

Notice that for 1 mol of KOH there is going to be 1 mol of OH. As well as in the RbOH. This means that for every 98.16 mol of KOH there are going to be 98.16 moles of OH-. Using the same logic, there is going to be 224.64 mol of OH- in the RbOH as well.

Total OH- = 98.16 + 224.64 = 322.8 mol

pOH = -log(322.8) = -2.51