Question

Given,50g of sample of sodium hydroxide required for complete neutralization, 1 litre 1N HCl. What is the percentage of purity of NaOH?
(A) 50
(B) 60
(C) 70
(D) 80

Answer 1

Hint: As we know that when acid reacts with base, neutralization reaction takes place. The question mentions complete neutralization which means that the number of moles of acid is equal to the number of moles of base. We will use this information in order to solve this question.
Formula used:
We will use the following formulas in this solution:-
Moles = $n-factor\times Normality\times Volume$
$n=\dfrac{\text{ mass}}{\text{molar mass}}$
Percentage purity = $\dfrac{\text{mass of pure compound in sample}}{\text{total mass of impure sample}}\times 100$

Complete answer:
Let us discuss about neutralization reaction followed by further calculations as follows:-
Neutralization reaction: When acid reacts with base, they neutralize each other and form salt and water molecules as products. This is referred to as the neutralization reaction. When sodium hydroxide (NaOH) reacts with (HCl), neutralization reaction takes place as shown below:-
$NaOH+HCl\to NaCl+{{H}_{2}}O$
-Calculation of moles of base NaOH:-
Moles of acid = $n-factor\times Normality\times Volume$
n-factor of HCl is 1.
The values given are as follows:-
Normality of HCl = 1N
Volume of HCl = 1litre
Moles of acid = $1\times 1\times 1=1moles$
As we know that for a complete neutralization, moles of acid will be equal to moles of base.
Therefore moles of base (n) = 1 mole
-Calculation of mass of NaOH which is purely present in sample:-
Molar mass of NaOH = (23 +16 +1) g/mol
$n=\dfrac{\text{ mass}}{\text{molar mass}}$
Rearrange the formula for calculation as follows:-
$\begin{align}
  & \Rightarrow \text{mass}=n\times \text{molar mass} \\
 & \Rightarrow \text{mass}=1mole\times 40g/mol \\
 & \Rightarrow \text{mass}=40g \\
\end{align}$
-Calculation of percentage purity of NaOH:-
Total sample of sodium hydroxide (NaOH) used = 50g
Percentage purity =$\dfrac{\text{mass of pure compound in sample}}{\text{total mass of impure sample}}\times 100$
Percentage purity of NaOH = $\dfrac{40}{50}\times 100=80%$

Therefore the percentage purity of NaOH is: (D) 80.

Note:
-Generally the n-factor of acids and bases are calculated by checking the number of hydrogen ions or hydroxide ions present in them respectively.
-Also use the values in calculations along with their units for better accuracy with less chance of errors.