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# 50 g of a sample of $\text{NaOH}$ required for complete neutralization of 1N $\text{HCl}$. What is the percentage purity of $\text{NaOH}$?

Last updated date: 04th Aug 2024
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Hint: To find the purity of $\text{NaOH}$, the equivalents of $\text{NaOH}$ need to be calculated. After finding the equivalents of $\text{NaOH}$ the percentage purity can be directly calculated by the formula: $\frac{\text{theoretical weight}}{\text{sample weight}}\times 100$

The question that comes to our minds is ‘What is percentage purity’? Percentage purity is the percentage of material actually useful or desired that is present in the sample. It can be calculated by the formula : $\frac{\text{weight of pure compound in sample}}{\text{total weight of impure sample}}\times 100$.

Step (1)- Find the equivalents- Let us calculate the mass of pure $\text{NaOH}$ in the sample. It will be calculated by calculating the equivalents of $\text{NaOH}$ that are required for neutralization of $\text{NaOH}$ with $\text{HCl}$. The reaction of neutralization of $\text{HCl}$ with $\text{NaOH}$ is $\text{NaOH}+\text{HCl}\to \text{NaCl}+{{\text{H}}_{2}}\text{O}$.

Step (2)- Calculate Molar masses- The reaction of neutralization made it clear that the moles of $\text{NaOH}$ and moles of $\text{HCl}$ required for the complete neutralization are the same. That is 1 mole of $\text{NaOH}$ requires 1 mole of $\text{HCl}$ in the neutralization process. 1 mole of $\text{NaOH}$ weighs
- Atomic mass of $\text{Na}$ is 23 grams.
- Atomic mass of $\text{O}$ is 16 grams.
- Atomic mass of $\text{H}$ is 1 gram. The molecular mass of $\text{NaOH}$ is 23+16+1 is equal to 40 grams. So, 1 mole of $\text{NaOH}$ weighs 40 grams. Thus, 40 grams of $\text{NaOH}$ is required for complete neutralization.

Step (3)- 1N of means 1M of $\text{HCl}$ because the n-factor of $\text{HCl}$ is 1. So, the equivalent mass of $\text{HCl}$ is equal to its molar mass similarly, with $\text{NaOH}$ the n-factor is 1. So, the molar mass will be equal to equivalent mass. 1 equivalent of $\text{NaOH}$ is equal to 40 grams of $\text{NaOH}$ which has taken part in the reaction. The mass of pure $\text{NaOH}$ that reacted with $\text{HCl}$ in the reaction is 40 grams.

Step (4)- Applying the formula of percentage purity, we will find the percentage of how much $\text{NaOH}$ is pure.
We are given that the total impure mass of $\text{NaOH}$ in the sample is 50 grams. Apply the formula;
Percentage purity= $\frac{\text{weight of pure compound in sample}}{\text{total weight of impure sample}}\times 100$
Substitute the required values in the formula, we will get,
Percentage purity=$\frac{40}{50}\times 100$, the zero will be cancelled the expression is now, $\frac{400}{5}$, which will be cancelled by 5. The purity percentage comes out to be 80%.
Hence, the Percentage purity of $\text{NaOH}$ in the sample is 80%.

Note:
Do not misunderstand the term percentage purity with yield percentage, as the formula is almost similar. But percentage purity is for finding the purity of a substance in the sample. But in yield percentage, how much product is obtained that needs to be found out.

Yield percentage= $\frac{\text{actual yield}}{\text{theoretical yield}}\times 100$ whereas Percentage purity= $\frac{\text{weight of pure compound in sample}}{\text{total weight of impure sample}}\times 100$