Question

g 304 mL of a 0.36 M potassium hydroxide solution is added to 341 mL of a 0.51 M lithium hydroxide solution. Calculate the pOH of the resulting solution.

Answer 1

Answer:

• pOH = 0.36

Explanation:

Both potassium hydroxide and lithium hydroxide solutions are strong bases, so you assume 100% dissociation.

1. Potassium hydroxide solution, KOH

• Volume, V = 304 mL = 0.304 liter

• Molarity, M = 0.36 M

• number of moles, n = M × V = 0.36M × 0.304 liter = 0.10944 mol

• 1 mole of KOH produces 1 mol of OH⁻ ion, thus the number of moles of OH⁻ is 0.10944

2. LIthium hydroxide, LiOH

• Volume, V = 341 mL = 0.341 liter

• Molarity, M = 0.51 M

• number of moles, n = M × V = 0.341 liter × 0.51 M = 0.17391 mol

• 1mole of LiOH produces 1 mol of OH⁻ ion, thus the number of moles of OH⁻ is 0.17391

3. Resulting solution

• Number of moles of OH⁻ ions = 0.10944 mol + 0.17391 mol = 0.28335 mol

• Volume of solution = 0.304 liter + 0.341 liter = 0.645 liter

• Molar concentration = 0.28335 mol / 0.645 liter = 0.4393 M

4. pOH

      $pOH=-\log [OH^-]=-\log (0.439)=0.36$   ← answer