Question

0.688g of antacid was treated with 50.00mL of 0.100M HCl. The excess acid required 4.21 mL of 1.200M NaOH for back titration. What is the neutralizing power of this antacid expressed as millimoles of HCl per gram of antacid?

Answer 1

We can solve this problem through the following steps

The moles of HCl treated with the antacid for neutralisation = M1xV1 = 50.00x10^-3 Lx0.100molL^-1 = 5x10^-3 mol

the moles of excess HCl that is not neutralised by the antacid = moles of NaOH required for back titration = M2xV2 = 4.21x10^-3 Lx1.200 molL^-1= 5.052x10^-3 mol

However this is not possible because the moles of excess HCl that is not neutralised by the antacid can never be greater than the initial moles of HCl taken.

So please recheck the numerical values of the question.