The Foundations of Chemistry

_?_ mol HCl10.2 mmol NaOH10.2 mmol HCl

We know the volume of the HCl solution, so we can calculate its molarity.

0.278 MHCl

EXAMPLE 11-6 Titration

A 43.2-mL sample of 0.236 Msodium hydroxide solution reacts completely with 36.7 mL of
a sulfuric acid solution. What is the molarity of the H 2 SO 4 solution?

H 2 SO 4 
2NaOH88nNa 2 SO 4 
2H 2 O

Plan

The balanced equation tells us that the reaction ratio is one millimole of H 2 SO 4 to two
millimoles of NaOH, which gives the unit factor, 1 mmol H 2 SO 4 /2 mmol NaOH.

H 2 SO 4 
2NaOH88nNa 2 SO 4 
2H 2 O

1 mmol 2 mmol 1 mmol 2 mmol

First we find the number of millimoles of NaOH. The reaction ratio is one millimole of H 2 SO 4
to two millimoles of NaOH, so the number of millimoles of H 2 SO 4 must be one half of the
number of millimoles of NaOH. Then we can calculate the molarity of the H 2 SO 4 solution
because we know its volume.

Solution

The volume of a solution (in milliliters) multiplied by its molarity gives the number of
millimoles of solute.

_?_ mmol NaOH43.2 mL NaOH soln10.2 mmol NaOH

0.236 mmol NaOH



1 mL NaOH soln

10.2 mmol HCl



36.7 mL HCl soln

_?_ mmol HCl



mL HCl soln

1 mmol HCl



1 mmol NaOH

The indicator phenolphthalein

changes from colorless, its color in

acidic solutions, to pink, its color in

basic solutions, when the reaction in

Example 11-5 reaches completion.

Note the first appearance of a faint

pink coloration in the middle beaker;

this signals that the end point is

near.