How many milliliters of 0.162 M HCI are needed to neutralize 25.0 mL of 0.076 M Ca(OH)2.?

Question

2HCI + Ca(OH)2 ------------ 2H2O + CaCI2
(not sure if I have the correct balance equation.)

Answer 1

Ok, to neutralize a solution, you need to equal :

N1*V1 = N2*V2

V1 and V2 = volumes

N1 and N2 = normality

N = M*(theta)

in the case of Ca(OH)2, theta = 2

in the case of HCL, theta = 1

0.162*V1 = 25*0.076*2

V1 = 23.15 mL

Hope that helps