A solution is made by dissolving 8.2 g of NaOH in approximately 450 mL of water in a volumetric flask. The solution becomes quite warm, but after it is allowed to return to room temperature, water is added to total 500 mL of solution. Calculate the pH of the final solution. Report pH to 2 decimal places.

13.61

Explanation:

First, we will calculate the molar concentration of NaOH in the final solution.

M = mass NaOH / molar mass NaOH × liters of solution

M = 8.2 g / 40.00 g/mol × 0.500 L

M = 0.41 M

NaOH is a strong base that dissociates according to the following balanced equation.

NaOH → Na⁺ + OH⁻

Then, the concentration of OH⁻ is also 0.41 M.

The pOH is:

pOH = -log [OH⁻]

pOH = -log 0.41 = 0.39

The pH is:

pH + pOH = 14

pH = 14 - pOH = 14 - 0.39 = 13.61

first, consult an activity series chart to determine if al is more reactive than fe. it is, so it will replace the fe. now, one can continue with the problem.

next, make sure you have a correctly balanced equation.

3feo + 2al ==> 3fe + al2o3

now, determine which reactant (feo or al) is in limiting supply.

moles feo present = 125 g x 1 mole/71.8 g = 1.74 moles

moles al present = 25.0 g x 1 mole/26.98 g = 0.927 moles

since from the balanced equation, it take 3 mol feo for every 2 mol al, al will be limiting.

you can determine this simply by dividing moles of each by their coefficient in the balanced equation. for example: 1.74/3 = 0.58 and 0.927/2 = 0.464. al is less, so it is limiting.

finally, using the stoichiometry in the balanced eq. determine mass of fe that can be produced from 0.927 moles of al.

0.927 moles al x 3 moles fe/2 moles al x 55.8 g/mole = 77.6 g