Chemistry

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

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asked by Caleb
  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

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