Answer to Question #115601 in Organic Chemistry for Arda

Question

What volume (mL) of 0.224 M NaOH is required to neutralize 25.1 mL of 0.388 M HCl?

A)43.5

B)0.0230

C)14.5

D)0.002180

E)2.180

Answer 1

NaOH + HCl = NaCl + H2O

number of moles (n = C*V, i.e. concentration * volume) of NaOH and HCl is equal, therefore:

C(NaOH)*V(NaOH) = C(HCl)*V(HCl)

V(NaOH) = C(HCl)*V(HCl)/C(NaOH) = 0.388 M * 25.1 ml / 0.224 M = 43.5 ml

43.5 ml of 0.224 M NaOH is required