Question

How many mL of a 0.81 M sodium hydroxide solution are required to react completely with 74 mL of a 0.12 M hydrochloric acid solution? Report your answer to the nearest mL.

Answer 1

The reaction between HCl and NaOH is as follows,

HCl + NaOH -----> NaCl + H2O

We see that, according to the stoichiometry of the reaction 1 mole of HCl reacts with 1 mole of NaOH

Given,

[NaOH] = 0.81 M

[HCl] = 0.12 M

Volume of HCl solution = 74 mL = 0.074 L

We know that,

Moles = Molarity x Volume (L)

=> Moles of HCl = 0.12 x 0.074 = 0.00888 moles

Since, Moles of HCl = Moles of NaOH

Moles of NaOH = 0.00888

=> 0.00888 = 0.81 x V

=> V = 0.01096 L = 10.96 mL = Volume of NaOH required