Determine the pOH of the solution.

Question

Determine the pOH (to two decimal places) of the solution that is produced by mixing 54.3 mL of 2.54×10^-3 M Na₂O with 923 mL of 7.45×10^-1 M MgO.

Answer 1

Write the balanced equation for the reaction taking place when placed in water:

Na₂O + H₂O ==> 2NaOH

MgO + H₂O ==> Mg(OH)₂

moles Na₂O = 54.3 ml x 1 L / 1000 ml x 2.54x10^-3 mol/L = 1.379x10^-4 moles

moles NaOH formed = 1.379x10^-4 moles Na₂O x 2 moles NaOH / mol Na₂O = 2.758x10^-4 mols NaOH

moles MgO = 923 ml x 1 L / 1000 ml x 7.45x10^-1 mol/L = 0.688 mols MgO

moles Mg(OH)₂ formed = 0.688 mols MgO x 1 mol Mg(OH)₂ / mol MgO = 0.688 mols Mg(OH)₂

Total moles OH^- produced:

From NaOH: 2.758x10^-4 mols

From Mg(OH)₂: 0.688 x 2 = 1.376 mols

TOTAL = 1.376 moles OH^-

Final volume = 54.3 ml + 923 ml = 977.3 mls = 0.9773 L

Final [OH-] = 1.376 mols OH^- / 0.9773 L = 1.408 M

pOH = -log [OH^-] = -log 1.408

pOH = 0.15