Question

What volume of 0.32M HCl is required to completely neutralize 50.0m mL of a 0.12M Mg(OH)2 solution?​

Answer 1

Answer:

37.5 mL of HCl will be required for complete neutralization with magnesium hydroxide

Explanation:

${ \bf{2HCl _{(aq)} + Mg(OH)_{2(s)}→ MgCl_{2(s)} + 2H _{2} O_{(l)}}}$

we've to first get moles of magnesium hydroxide in 50.0 ml :

${ \sf{1 \: l \: of \: hydroxide \: contains \: 0.12 \: moles}} \\ { \sf{0.05 \: l \: of \: hydroxide \: contain \: (0.05 \times 0.12) \: moles}} \\ = { \underline{0.006 \: moles}}$

for complete neutralization:

${ \sf{1 \: moles \: of \: hydroxide\: react \: with \: 2 \: moles \: of \: acid}} \\ { \sf{0.006 \: moles \: react \: with \: \{0.006 \times 2 \}} \: moles} \\ = { \underline{0.012 \: moles \: of \: acid}}$

compare with acid molarity:

${ \sf{0.32 \: moles \: of \: acid \: occupy \: 1 \: litre}} \\ { \sf{0.012 \: moles \: of \: acid \: occupy \: ( \frac{0.012}{0.32}) \: litres }} \\ { \underline{ = 0.0375 \: litre \: \: = \: \: 37.5 \: ml}}$