Find the pOH of the following mixture

Question

determine the pOH of the solution that is the produced by mixing 247 mL of 1.27x10^-1 M of MgH₂ with 4.99 mL of 4.60x10^-4 M Ca(OH)₂

Answer 1

MgH₂ is magnesium hydride, which when placed in water produces Mg(OH)_2, magnesium hydroxide.

MgH₂ + 2 H₂O ==> Mg(OH)₂ + H₂

moles MgH₂ = 0.247 L x 0.127 mol/L = 0.03137 moles

moles OH^- = 0.03137 mol MgH₂ x 2 mol OH^-/mol MgH₂ = 0.0627 = 6.27x10^-2 moles OH

moles Ca(OH)₂ = 0.00499 L x 4.6x10^-4 mol/L = 2.295x10^-6 mol

moles OH^- = 2.295x10^-6 mol Ca(OH)₂ x 2 = 4.59x10^-6 mol OH

Total moles OH^- = 6.27x10^-2 + 4.59x10^-6 = 6.27x10^-2 moles OH^-

Total volume = 247 ml + 4.99 ml = 251.99 ml = 0.252 L

Final [OH^-] = 6.27x10^-2 mol OH^- / 0.252 L = 0.249 M

pOH = -log [OH^-] = - log 0.249

pOH = 0.60