In a titration of 0.608 L of 1.222 M B solution, how many liters of 1.088 M A solution are required to reach the end-point.

Question

A + 2B → 2C + D

Answer 1

Looking at the balanced equation...

A + 2B ==> 2C + D, we see that it takes 1 mole of A to neutralize (react with) 2 moles of B. So, if we know the moles of B present (which we can find from volume and molarity), we can then find mole of A needed. Once we have moles of A needed, we can determine the volume (see below).

moles of B present = 0.608 L x 1.222 moles/L = 0.743 moles

moles of A needed = 0.743 moles B x 1 mole A / 2 moles B = 0.372 moles of A needed

Since A is 1.088 M, we know it has 1.088 moles A/liter. Thus,...

(x L)(1.088 mol/L) = 0.372 moles

x = 0.342 liters of A are required