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.

First, we have to calculate the concentration of each compound 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

After this, we are going to use the density of each compound to found their mass and then their molar mass to find their moles. For KOH:

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

For RbOH

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

OH - Moles

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

pOH = - log (322.8) = - 2.51