Question

Instead of using ratios for back titrations we can also use molarities, if our solutions are standardized. A 0.196 g sample of antacid containing an unknown amount of triprotic base Al(OH)3 was reacted with 25.0mL of 0.111M HCl. The resulting solution was then titrated with 11.05mL of 0.132M NaOH solution. Calculate the mass percent of Al(OH)3 in the antacid sample.

Answer 1

Answer:

The mass percent of Al(OH)₃ is 15.3%

Explanation:

The reaction is:

Al(OH)₃ + 3HCl = AlCl₃ + 3H₂O

The excess acid is neutralized with a solution of sodium hidroxide, in the reaction:

NaOH + HCl = NaCl + H₂O

The total moles of HCl is:

$n_{HCl,total} =M_{HCl} *V_{HCl} =0.111*0.025=2.78x10^{-3} moles$

From the second titration, the moles of excess of HCl is:

$n_{HCl,excess} =n_{NaOH} =M_{NaOH} *V_{NaOH} =0.132*0.01105=1.46x10^{-3} moles$

The difference between the total and excess of HCl, it can be know the moles that reacts with the aluminum hydroxide, is:

$n_{HCl,reacts} =n_{HCl,total}-n_{HCl,excess} =2.78x10^{-3} moles-1.46x10^{-3} moles=1.32x10^{-3} moles$

The ratio between HCl and Al(OH)₃ is 3:1. The MW for aluminum hydroxide is 78 g/mol, thus:

$m_{Al(OH)3} =1.32x10^{-3} molesHCl*\frac{1molAl(OH)3}{3molesHCl} *\frac{78gAl(OH)3}{1molAl(OH)3} =0.03g$

The percentage of Al(OH)₃ is:

$Percentage-Al(OH)3=\frac{m_{Al(OH)3} }{m_{antiacid} } *100=\frac{0.03}{0.196} =15.3$%