Question

What volume of $0.20M$ $Ca{\left( {OH} \right)_2}$ will neutralize $45.0{\text{ }}mL$ of a $1M$ solution of $HCl{O_3}$?

Answer 1

A neutralization reaction can be defined as a chemical reaction in which an acid and base quantitatively react together to form a salt and water as products. In a neutralization reaction, there is a combination of ${H^ + }$ ions and $O{H^ - }$ ions which form water. A neutralisation reaction is generally an acid-base neutralization reaction.
$Acid + Base \to {H_2}O + Salt$
When a solution is neutralized, it means that salts are formed from equal weights of acid and base. The amount of acid needed is the amount that would give one mole of protons (${H^ + }$) and the amount of base needed is the amount that would give one mole of ($O{H^ - }$). Because salts are formed from neutralization reactions with equivalent concentrations of weights of acids and bases: N parts of acid will always neutralize N parts of base.

Complete answer:
Formula to calculate number of moles is given by:

$$Molarity,M = \dfrac{{Number{\text{ }}of{\text{ }}moles,n}}{{Volume{\text{ }}of{\text{ }}solution,V\left( {in{\text{ }}L} \right)}} \\\ n = M \times V \\\$$

Now, for given acid and base neutralization reaction is as follows:
$Ca{\left( {OH} \right)_2}\; + \;2HCl{O_3}\; \to \;Ca{\left( {Cl{O_3}} \right)_2}\; + \;2{H_2}O$
For neutralisation reaction to take place,
Number of equivalents of$Ca{\left( {OH} \right)_2}$ = Number of equivalents of $HCl{O_3}$
$Number{\text{ }}of{\text{ }}equivalents{\text{ }} = n \times n.f$
Where, $n$= number of moles
$n.f$= n factor (which is acidity and basicity in case of acids and bases)
Here, $n.f$ of $Ca{\left( {OH} \right)_2}$ = $2$
$n.f$ of $HCl{O_3}$ = $1$

we know, $Molarity,M = \dfrac{{Number{\text{ }}of{\text{ }}moles,n}}{{Volume{\text{ }}of{\text{ }}solution,V\left( {in{\text{ }}L} \right)}}$
$n = M \times V$
Given, M of $HCl{O_3}$,${M_1} = 1M$
V of $HCl{O_3}$, ${V_1} = 45mL$
M of $Ca{\left( {OH} \right)_2}$, ${M_2} = 0.20M$
V of $Ca{\left( {OH} \right)_2}$,${V_2} = {V_2}mL$
Now, putting the values in number of equivalents formula
${n_1} \times n.{f_1} = {n_2} \times n.{f_2}$

$$1 \times 45 \times 1 = 0.20 \times {V_2} \times 2 \\\ {V_2} = \dfrac{{45}}{{2 \times 0.20}} \\\ {V_2} = 112.5mL \\\$$

Hence, $112.5mL$of $0.20M$ $Ca{\left( {OH} \right)_2}$ will neutralize $45.0{\text{ }}mL$ of a $1M$ solution of $HCl{O_3}$.

Additional Information:
The neutralization of a strong acid and strong base has a pH equal to 7. The neutralization of a strong acid and weak base will have a pH of less than 7, and conversely, the resulting pH when a strong base neutralizes a weak acid will be greater than 7.
When an acid is neutralized the amount of base added to it must be equal the amount of acid present initially. This amount of base is said to be the equivalent amount. In a titration of an acid with a base, the point of neutralization can also be called the equivalence point.

Note:
For acids, n-factor is defined as the number of ${H^ + }$ ions replaced by 1 mole of acid in a reaction. The n-factor for acid is not equal to its basicity; i.e. the number of moles of replaceable ${H^ + }$ atoms present in one mole of acid. Bases are the species, which furnish $O{H^ - }$ ions when dissolved in a solvent. For bases, n-factor is defined as the number of $O{H^ - }$ ions replaced by 1 mole of base in a reaction. n-factor is not equal to its acidity i.e. the number of moles of replaceable $O{H^ - }$ ions present in 1 mole of base.