In a titration, 25.9 mL of 3.4 x10^-3 M Ba (OH) 2 neutralized 16.6 mL of HCL solution. What is the molarity of the HCL solution?

First, We have to write the equation for neutralization:

Ba (OH) 2 + 2HCl → BaCl2 + 2H2O

so, from the equation of neutralization, we can get the ratio between Ba (OH) 2 and HCl. Ba (OH) 2 : HCl = 1:2

- We have to get the no. of moles of Ba (OH) 2 to do the neutralization as we have 25.9ml of 3.4 x 10^-3 M Ba (OH) 2.

So no. of moles of Ba (OH) 2 = (25.9ml/1000) * 3.4x10^-3 = 8.8 x 10^-5 mol

and when Ba (OH) 2 : HCl = 1: 2

So the no. of moles of HCl = 2 * (8.8x10^-5) = 1.76 x 10^-4 mol

So when we have 1.76X10^-4 Mol in 16.6 ml (and we need to get it per liter)

∴ the molarity = no. of moles / mass weight

= (1.76 x 10^-4 / 16.6ml) * (1000ml/L) = 0.0106 M Hcl