20 mL of 0.02 M sodium hydroxide was required to neutralize 20 mL of 0.10 M hydrochloric acid to which 1 gram of Milk of Magnesia...?

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2018-02-14 1:43 pm

3) In a “back titration” experiment, 20 mL of 0.02 M sodium hydroxide was required to neutralize 20 mL of 0.10 M hydrochloric acid to which 1 gram of Milk of Magnesia tablet was added prior to titration. Calculate the following. Show all work.
a) Calculate the moles of magnesium hydroxide in the 1- gram tablet. b) Calculate the mass of magnesium hydroxide in the 1- gram tablet. c) Calculate the percent active ingredient in the 1- gram tablet of Milk of Magnesia.

回答 (1)

wanszeto

2018-02-14 2:50 pm

3)
a)
The titration of excess hydrochloric acid against standard sodium hydroxide :
HCl + NaOH → NaCl + H₂O
Mole ratio HCl : NaOH = 1 : 1

No. of moles of NaOH used in titration = (0.02 mol/L) × (20/1000 L) = 0.0004 mol
No. of moles of excess HCl = 0.0004 mol

Total no. of moles of HCl added = (0.10 mol/L) × (20/1000 L) = 0.0020 mol
No. of moles of HCl reacted with Mg(OH)₂ = (0.002 - 0.0004) mol = 0.0016 mol

The reaction of hydrochloric acid with magnesium hydroxide in Milk of Magnesia tablet :
2HCl + Mg(OH)₂ → MgCl₂ + 2H₂O
Mole ratio HCl : Mg(OH)₂ = 2 : 1

No. of moles HCl reacted with Mg(OH)₂ = 0.0016 mol
No. of moles of Mg(OH)₂ in 1-gram of Milk of Magnesia tablet = (0.0016 mol) × (1/2) = 0.00080 mol (to 2 sig. fig.)

b)
Molar mass of Mg(OH)₂ = (24.3 + 16.0×2 + 1.0×2) g/mol = 58.3 g/mol
Mass of Mg(OH)₂ in 1-gram of Milk of Magnesia tablet = (0.00080 mol) × (58.3 g/mol) = 0.047 g (to 2 sig. fig.)

c)
Percent active ingredient = (0.047/1) × 100(%) = 4.7% (to 2 sig. fig.)

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