Question

It required 28.9 mL of 0.0170-M Ba(OH)2 solution to titrate a 25.0-mL sample of HCl to the equialence point. Calculate the molarity of the HCl solution. Molarity = M

Answer 1

Moles of Ba(OH)₂ required = 28.9 mL x 0.0170 M = 0.491 mmol

Let us say that the molarity of HCl = Z M

Hence, the moles of HCl titrated = 25.0 mL x Z M = 25.0Z mmol

2HCl(aq) + Ba(OH)₂(aq) $\rightarrow$ BaCl₂(aq) + 2H₂O(l)

Now, 2 mol of HCl is titrated by 1 mol of Ba(OH)₂

Therefore, 25.0Z mmol of HCl is titrated by = 25.0Z mmol x (1 mol/2 mol) = 12.5Z mmol of Ba(OH)₂

Thus,

12.5Z mmol = 0.491 mmol

or, Z = 0.0393 M

Hence, the molarity of the HCl solution = 0.0393 M