Question

A titration is carried out between hydrochloric acid and sodium hydroxide. At the beginning of the experiment, 12.00 mL of 0.9881 M hydrochloric acid is placed in an Erlenmeyer flask with additional water and the indicator, phenolphthalein. What is the molarity of the sodium hydroxide solution if it takes 5.55 mL to neutralize the acid and reach the end point? In the flask below, draw the chemical substances (other than water) present at the end point of the titration.

Answer 1

The reaction of HCl and NaOH is as follows:

HCl + NaOH = NaCl + H2O

The chemical substances (other than water) present at the end point of the titration are , soium and chloride ions

NaCl (aq) = Na+ (aq) + Cl –(aq)

Given that molarity of HCl = 0.9881 M

Volume of HCl ; 12.00 ml or 0.012 L

Volume of NaOH = 5.55 ml or 0.00555L

First calculate the number of moles of HCl as follows:

Number of mole = molarity * volume in L

= 0.9881 M *0.012 L

= 0.012 mole HCl

Now calculate the mole of NaOH as follows:

0.012 mole HCl * 1/1

= 0.012 mole NaOH

Molarity = number of moles / volume in L

= 0.012 mole NaOH /0.00555L

= 2.16 M NaOH