Question

How many moles of $ {\text{KOH}} $ are needed to exactly neutralize $ 500 {\text{mL}} $ of $ 1.0{\text{M}} {\text{HCl}} $ ?

Answer 1

Hint: This problem can easily be solved with the help of basic stoichiometry and fundamentals of concentration of liquids. First, we will compare the reactant side with the products and then we will proceed after balancing the chemical equation. We shall calculate the number of moles of hydrogen present which will be equal to the moles of KOH required.

Formula Used
We would require the formula for molarity to solve this question
 $ {\text{molarity}} = \dfrac{{{n_s}}}{{{v_s}}} $
Where
 $ {n_s} $ is the number of moles of solute
 $ {v_s} $ is the volume of solvent in litres.

Complete Step-by-Step Solution
According to the question, the following information is provided to us
Molarity of $ HCl $ solution $ = 1 {\text{M}} $
Volume of $ HCl $ solution $ = 500 {\text{mL}} = 0.5{\text{L}} $
In the above stated reaction potassium hydroxide, $ {\text{KOH}} $ , will react with hydrochloric acid, $ {\text{HCl}} $ , to produce aqueous potassium chloride and water.
The neutralization reaction happens between an acid and a base and it generally forms salt and water. The above-mentioned reaction can be written below.
 $ {\text{KO}}{{\text{H}}_{(aq)}} + {\text{HC}}{{\text{l}}_{(aq)}} \to {\text{KO}}{{\text{H}}_{(aq)}} + {{\text{H}}_2}{{\text{O}}_{(l)}} $
Therefore, the reaction consumes potassium hydroxide and hydrochloric acid at a molar ratio of 1: 1, which means that the number of moles of potassium hydroxide required to completely neutralize the solution of hydrochloric acid is equal to the number of moles of hydrochloric acid present in the solution.
We know that molarity tells us the number of moles of solute per unit volume of solvent (in litres). Now, we will use the above formula to get the result.
 $ {\text{molarity}} = \dfrac{{{n_s}}}{{{v_s}}} $
We can rewrite it as
 $ {n_s} = {\text{molarity}} \times {v_s} $
Upon substituting values, we get
 $ {n_s} = 1M \times 0.5L $
Upon solving, we get $ {n_s} = 0.5 $ moles of $ {\text{KOH}} $
Hence, we require $ 0.5 $ moles of $ {\text{KOH}} $ to exactly neutralize $ 500 {\text{mL}} $ of $ 1.0{\text{M}} {\text{HCl}} $ .

Note
A neutralization reaction can be defined as a chemical reaction in which an acid and a base react together quantitatively to form a product of salt and water.
There is a combination of $ {H^ + } $ ions and $ O{H^ - } $ ions which form water in a neutralization reaction. In general, a neutralisation reaction is an acid-base neutralization reaction.