Knowing that the milk of magnesia contains 8 g% of Mg (OH)2, using 1 g of milk of magnesia and added 50 mL of 0.100 M HCl, determine the required volume of a solution of 0.105 M NaOH to titrate the excess HCl 0.100 M.

Answer 1

1-g of MM contains 8 wt% Mg(OH)2 (f.wt=58.33g/mole)

8% of 1gm = 0.08(1)g Mg(OH)2
= 0.08g Mg(OH)2
= (0.08/58.33)mole Mg(OH)2
= 0.00137 mole Mg(OH)2

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

Adding 50ml of 0.100M HCl
= 0.050(0.10)mole HCl
= 0.005mole HCl

0.005mole HCl + 0.00274mole OH^-
=> (0.005 - 0.00274)mole excess HCl
=> 0.0023mole excess HCl
=> 0.0023mole excess H^+

Concentration of excess (H^+) in 50ml solution:
[H^+] = (0.0023/0.050)M = 0.045M

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

Answer 2

Thanks :)

Answer 3

You are welcome.