Question

When a chemist titrates a standard solution of 168.61 mL of hydrochloric acid (HCl) with 0.104 M sodium hydroxide (NaOH) , she finds that it requires 259.4 mL of the base to reach the endpoint of the titration. What is the molarity of the acid solution ?

Answer 1

The conc of NaOH is 0.104M which means that 1000mL solution contains 0.104mol of NaOH, so the amount of moles used in the titration will be

`(0.104mol)/(1000mL) * 259.4mL = 4.00* 10^(-7) mol`
The reaction of NaOH and HCl is one to one so the moles of HCl that were titrated will also be
`4.00* 10^(-7) mol`
The Molarity is therefore

`(4.00* 10^(-7) mol)/(168.61 mL) * 1000mL = 2.38* 10^(-6) M`