Question

milk of magnesia is a common liquid antacid that contains magnesium hydroxide. since magnesium hydroxide isnt highly soluble in water milk of magnesia is sold as a suspension (thick liquid) where the solute isnt dissolve it stays as a small solid particles homogeneously mixed throughout the liquid; it is estimated that 1.00 ml of milk of magnesia contains 80.0 mg of magnesium hydroxide . assuming that our stomach acid has a concentration of 0.166M HCL how much milk of magnesia is needed to neutralize 100ml of our stomach acid?

Answer 1

Consider a neutralization reaction,  2 HCl + Mg(OH) ₂ MgCl ₂+ 2 H₂O

From reaction, stoichiometric ratio = No of moles of acid / No of moles of base = 2/1=2

We have correlation, No. of moles of acid = No. of moles of base x stoichiometric ratio.

We have, Molarity = No. of moles of solute / Volume of solution in L

$\therefore$ No. of moles of HCl = Molarity $\times$ Volume of solution in L = 0.166 mol / L $\times$ 0.100 L = 0.0166 mol

We can write, 0.0166 mol HCl = No. of moles of Mg(OH) ₂$\times$ 2

No. of moles of Mg(OH) ₂ = 0.0166 mol / 2 = 8.3 $\times$ 10 ^-03 mol

Molar mass of Mg(OH) ₂ = 24.31 + ( 2*16.00) + (2*1.0079) = 58.32 g/mol

We have, No. of moles = Mass / Molar Mass

Therefore, Mass of Mg(OH) ₂ = No. of moles $\times$ Molar mass = 8.3 $\times$ 10 ^-03 mol $\times$ 58.32 g/mol = 0.484 g

Mass of Mg(OH) ₂ = 0.484 g = 484 mg

1 ml milk of magnesia contain 80 mg Mg(OH) ₂.

Therefore, Volume of milk of magnesia contain 484 mg Mg(OH) ₂.= 484 mg $\times$ 1 ml / 80 mg = 6.1 ml

ANSWER : 6.1 ml milk of magnesia will be required to neutralize 100 ml of our stomach acid.