Question

How much of NaOH is required to neutralize $\text{1500c}{{\text{m}}^{\text{3}}}$ of 0.1N HCl (Na=23)?
A) 40g
B) 4g
C) 6g
D) 60g

Answer 1

The molecular mass of NaOH is 40kg/mol
We use the equation, $\text{Weight}\,\text{of}\,\text{NaOH = Normality }\\!\\!\times\\!\\!\text{ Equivalent}\,\text{weight }\\!\\!\times\\!\\!\text{ }\dfrac{\text{Volume}}{\text{1000}}$

Complete Solution :
So here in the question it is asked to neutralize $\text{1500c}{{\text{m}}^{\text{3}}}$ of 0.1N HCl solution how much grams of NaOH required.
- We know that the NaOH is a strong base and HCl is a strong acid, both dissociates completely in the aqueous solution and yields a neutral salt and water as the final products.
- To solve this question, we should know that to undergo neutralization completely the molar equivalents of HCl and NaOH should be equal.
So we can write, molar equivalents of NaOH = molar equivalents of HCl
${{M}_{eq}}(NaOH)={{M}_{eq}}\left( HCl \right)$
First let’s write the data given in the question.
Normality of HCl = 0.1N
Volume of HCl = $\text{1500c}{{\text{m}}^{\text{3}}}$
And hence we can calculate molar equivalents of HCl as,
${{\text{M}}_{\text{eq}}}\left( \text{HCl} \right)\text{=Normality }\\!\\!\times\\!\\!\text{ Volume}$
${{\text{M}}_{\text{eq}}}\left( \text{HCl} \right)\text{=0}\text{.1}\times \text{1500}\times \text{1}{{\text{0}}^{-3}}$
${{\text{M}}_{\text{eq}}}\left( \text{HCl} \right)\text{=0}\text{.15}$
Since the volume is given in $c{{m}^{3}}$ which is equal to 1ml, we have to convert the ml value into L.
Now we have to find the weight of NaOH required for the neutralization of 0.15M of HCl.
We can find the value of weight of the NaOH using the equation,
${{\text{M}}_{\text{eq}}}\text{(NaOH)=}{{\text{M}}_{\text{eq}}}\text{(HCl)=}\dfrac{\text{Weight}\,\text{of}\,\text{NaOH}\,\text{required}}{\text{Equivalent}\,\text{weight}\,\text{of}\,\text{NaOH}}$
As we need the equivalent weight of NaOH.
Equivalent weight of NaOH can be calculated by dividing the molecular weight of NaOH and the n-factor.
n-factor is the number of replaceable $O{{H}^{-}}$ ions present in one mole of a base.
The molecular weight of NaOH is 40g and the n-factor is 1, since there is only one $O{{H}^{-}}$ion in one mole of NaOH that can be displaced during a reaction.
$\text{Equivalent}\,\text{weight=}\dfrac{\text{Molecular}\,\text{weight}}{\text{n-factor}}=\dfrac{40}{1}=40$
Before substituting the values the equation should be rearranged as,
$\text{Weight of NaOH required=}{{\text{M}}_{\text{eq}}}\text{(NaOH) }\\!\\!\times\\!\\!\text{ Equivalent}\,\text{weight}$
$\text{Weight of NaOH required=0}\text{.15 }\\!\\!\times\\!\\!\text{ 40=6g}$
Hence we got the amount of NaOH required for the neutralization reaction of HCl.
So, the correct answer is “Option C”.

Note: This question can also be directly solved by the direct equation,
$\text{Weight}\,\text{of}\,\text{NaOH = Normality }\\!\\!\times\\!\\!\text{ Equivalent}\,\text{weight }\\!\\!\times\\!\\!\text{ }\dfrac{\text{Volume}}{\text{1000}}$
- Here we know normality of HCl for $\text{1500c}{{\text{m}}^{\text{3}}}$ of the solution is 0.1N
Substitute all the values directly in the equation, as this equation is the altered form for the equation of one normal solution.
$\text{Weight}\,\text{of}\,\text{NaOH = 0}\text{.1 }\\!\\!\times\\!\\!\text{ 40 }\\!\\!\times\\!\\!\text{ }\dfrac{1500}{\text{1000}} = 6g$