Question

Calculate the number of milliliters of 0.738 M NaOH required to precipitate all of the Mg2+ ions in 148 mL of 0.541 M MgSO4 solution as Mg(OH)2. The equation for the reaction is:

MgSO4(aq) + 2 NaOH(aq) Mg(OH)2(s) + Na2SO4(aq)

_____________mL NaOH

Answer 1

How many mols MgSO4 so you have? That is moles = M x L = 0.541 x 0.148 = ?

Using the cofficients in the balanced equation, convert moles MgSO4 to moles NaOH. That is
?moles MgSO4 x (2 moles NaOH/1 mol MgSO4) = ?moles MgSO4 x 2/1 = ? moles MgSO4.

Then M = moles/L soln
MNaOH = moles NaoH/L NaOH
You know M NaOH and moles NaOH, solve for L NaOH and convert to mL.