Answer to Question #169855 in General Chemistry for Alexis

Question

How many mL of .145 of M HCL are needed to completely neutralize 49.0 mL of 0.100 M Ba(OH)2 solution

Answer 1

Reaction:

2HCl + Ba(OH)₂ = BaCl₂ + 2H₂O

n(Ba(OH)₂) = c * V = 0.1M * 0.049L = 0.0049mol

n(HCl) = 2n(Ba(OH)₂) = 0.0098mol

V(HCl) = n / c = 0.0098mol / 0.145M = 0.0676L = 67.6mL