What volume of 0.345 M HCl is required to neutralize 50.0 mL of 0.800 M NaOH?

Answer 1

HCl + NaOH ==> NaCl + H2O .. acid/base neutralization

moles NaOH present = 50.0 ml x 1 L/1000 ml x 0.800 moles/L = 0.0400 moles

It will take 0.0400 moles HCl to completely neutralize this (see mole ratio in balanced equation)

(x L)(0.345 mole/L) = 0.0400 moles

x = 0.1159 liters = 115.9 mls = 116 mls of HCl needed (3 sig. figs.)

Since the mole ratio of HCl : NaOH is 1 : 1 you can also solve this as follows:

(x ml)(0.345 m) = ( 50.0 ml)(0.800 M)

x = 115.9 mls = 116 mls (3 sig. figs.)