Knowing the milk of magnesia contains 8 g% of Mg (OH) 2, we used 1 g milk of magnesia and added 50 mL of 0.100 M HCl, determine the volumeainda of a solution of 0.105 M NaOH to titrate the excess HCl 0.100 M

Answer 1

Given 1gm of MM contains 8% Mg(OH)2 (f.wt. = 58.33 g/mol)

Gms Mg(OH)2
= 8% of 1gm MM = 0.08gms Mg(OH)2
= 0.00137mol Mg(OH)2

Mg(OH)2 in solution => 2moles OH^-
Moles OH^- = 2(0.00137)mole OH^-
= 0.00274mole OH^-

Adding 50ml of 0.100M HCl
=> 0.005mole HCl + 0.00274 mole OH^-
=>(0.0050 - 0.00274)mole excess(H^+)
=> 0.0023mole excess (H^+)
=> (0.0023/0.050)M H^+
=> 0.045M in excess H^+

Titrating 50ml of 0.045M H^+ with 0.105M NaOH
=>(M x V)acid = (M x V)base
=>(0.045M)(50ml)=(0.105M)(Vol Base)
=> Vol Base = (0.045M)(50ml)/(0.105M)
=> Vol Base = 21.4ml of 0.105M NaOH