Question

Calculate the number of milliliters of 0.687 M Ba(OH)₂ required to precipitate all of the Mg²⁺ ions in 196 mL of 0.499 M MgSO₄ solution as Mg(OH)₂.

Answer 1

Balanced equation : MgSO₄ (aq) + Ba(OH)₂ (aq) $\rightarrow$ Mg(OH)₂ (s) + BaSO₄ (s)

Concentration MgSO₄ = 0.499 M

volume of MgSO₄ solution = 196 mL

moles of MgSO₄ = (concentration MgSO₄) * (volume of MgSO₄ solution)

moles of MgSO₄ = (0.499 M) * (196 mL)

moles of MgSO₄ = 97.8 mmol

moles of Ba(OH)₂ required = moles of MgSO₄

moles of Ba(OH)₂ required = 97.8 mmol

volume of Ba(OH)₂ required = (moles of Ba(OH)₂ required) / (concentration of Ba(OH)₂)

volume of Ba(OH)₂ required = (97.8 mmol) / (0.687 M)

volume of Ba(OH)₂ required = 142.4 mL