Substituting the HCl information in the equation, we haveSince 1 mol of HCl neutralizes 1 mol of NaOH, 0.001 mol of NaOH must be
present in the mixture. Since 1 mol NaOH = 40.0 g
then
0.01 mol × 40.0 g/mol = 0.04 g NaOHTherefore, 0.04 g of NaOH was in the 0.100-g sample of the solid mixture.
The percent is 0.04 g/0.100 g × 100 = 40%.
In the explanations given to this point, the reactions that took place were
between monoprotic acids (single hydrogen ions) and monobasic bases (one
hydroxide ion per base). This means that each mole of acid had 1 mole of
hydrogen ions available, and each mole of base had 1 mole of hydroxide ions
available, to interact in the following reaction until the end point was reached:
H+(aq) + OH−(aq) → 2H 2 O(l)This is not always the case, however, and it is important to know how to
deal with acids and bases that have more than one hydrogen ion and more than
one hydroxide ion per formula. The following is an example of such a problem.
Example 3
If 20.0 milliliters of an aqueous solution of calcium hydroxide, Ca(OH) 2 , is used
in a titration, and an appropriate indicator is added to show the neutralization
point (end point), the few drops of indicator that are added can be ignored in the
volume considerations. Therefore, if 25.0 milliliters of standard 0.050 M HCl is
required to reach the end point, what was the original concentration of the
Ca(OH) 2 solution?
The balanced equation for the reaction gives the relationship between the
number of moles of acid reacting and the number of moles of base:
The mole relationship here is that the number of moles of acid is twice the
number of moles of base: