Question

you take an antacid tablet which weighs 3.56g and you add 5.00ml of a 4.00M solution. you then titrate the solution with .358 M NaOh. the Sample uses 22.35ml of NaOH. What mass of Al(OH)3 is in the pill?

Answer 1

Solution :-

Lets calculate the moles of the Acid

moles of acid = 4.00 mol per L * 0.005 L = 0.02 mol

moles of NaOH reacted = 0.358 mol per L * 0.02235 L = 0.008 mol

so the moles of acid reacted with the antacid = 0.02 mol - 0.008 mol = 0.012 mol

now lets calculate the moles of Al(OH)2

0.012 mol acid * 1 mol Al(OH)3 / 3 mol acid = 0.004 mol Al(OH)3

now lets calculate the moles of Al(OH)3 to its mass

mass = moles * molar mass

         = 0.004 mol *78.0036 g per mol

        = 0.312 g Al(OH)3

so the mass of Al(OH)3 in the pill is 0.312 g