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Mobile App Chapter 15: Problem 35
A \(0.200\)-g sample of impure \(\mathrm{NaOH}\) requires \(18.25 \mathrm{~mL}\) of \(0.2406\) \(M \mathrm{HCl}\) for neutralization. What is the percent of \(\mathrm{NaOH}\) in the sample?
Short Answer
Expert verified
87.82%
Step by step solution
01
- Write the balanced chemical equation
The balanced chemical equation for the reaction between \(\text{NaOH}\) and \(\text{HCl}\) is: \[ \text{NaOH} + \text{HCl} \rightarrow \text{NaCl} + \text{H}_2\text{O} \].
02
- Calculate moles of \(\text{HCl}\)
Use the molarity formula \[\text{Molarity} = \frac{\text{moles}}{\text{volume (L)}}\right\] to find the moles of \(\text{HCl}\). \[ \text{Moles of HCl} = 0.2406 \ \text{M} \times 0.01825 \ \text{L} = 0.00439095 \ \text{mol}\]
03
- Use mole ratio to find moles of \(\text{NaOH}\)
From the balanced equation, the mole ratio of \(\text{NaOH}\) to \(\text{HCl}\) is 1:1. Therefore, moles of \(\text{NaOH}\) = moles of \(\text{HCl}\) = 0.00439095 \(\text{mol}\).
04
- Calculate mass of \(\text{NaOH}\)
Use the molar mass of \(\text{NaOH}\), which is 40.00 \(\text{g/mol}\). \[\text{Mass of NaOH} = \text{moles of NaOH} \times \text{molar mass of NaOH} \] \[\text{Mass of NaOH} = 0.00439095 \ \text{mol} \times 40.00 \text{g/mol} = 0.175638 \text{g} \].
05
- Calculate percent purity of \(\text{NaOH}\)
Use the formula \[\text{Percent Purity} = \left(\frac{\text{mass of NaOH}}{\text{mass of sample}} \right) \times 100\] \[\text{Percent Purity} = \left(\frac{0.175638 \ \text{g}}{0.200 \ \text{g}} \right) \times 100 = 87.82\%\]
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Stoichiometry
Stoichiometry is the part of chemistry that deals with the quantitative relationships among reactants and products in a chemical reaction. In stoichiometric calculations, we use the balanced chemical equation to determine the mole ratio of reactants and products.
In our exercise, the balanced equation is: \(\text{NaOH} + \text{HCl} \rightarrow \text{NaCl} + \text{H}_2\text{O}\).
This equation tells us that one mole of \(\text{NaOH}\) reacts with one mole of \(\text{HCl}\) to produce one mole of \(\text{NaCl}\) and one mole of water. This 1:1 mole ratio helps to simplify the calculation of moles needed and produced in the reaction.
In our exercise, the balanced equation is: \(\text{NaOH} + \text{HCl} \rightarrow \text{NaCl} + \text{H}_2\text{O}\).
This equation tells us that one mole of \(\text{NaOH}\) reacts with one mole of \(\text{HCl}\) to produce one mole of \(\text{NaCl}\) and one mole of water. This 1:1 mole ratio helps to simplify the calculation of moles needed and produced in the reaction.
Neutralization Reaction
A neutralization reaction is a type of chemical reaction in which an acid reacts with a base to produce a salt and water.
The general form is: \(\text{acid} + \text{base} \rightarrow \text{salt} + \text{water}\).
In our case, \(\text{HCl}\), a strong acid, reacts with \(\text{NaOH}\), a strong base, to form \(\text{NaCl}\), a salt, and water. Such reactions are essential in titration experiments where we accurately measure the volume needed to reach the endpoint of the reaction.
The general form is: \(\text{acid} + \text{base} \rightarrow \text{salt} + \text{water}\).
In our case, \(\text{HCl}\), a strong acid, reacts with \(\text{NaOH}\), a strong base, to form \(\text{NaCl}\), a salt, and water. Such reactions are essential in titration experiments where we accurately measure the volume needed to reach the endpoint of the reaction.
Percent Purity Calculation
Percent purity shows how much of a sample's mass is the actual desired compound versus impurities. To find percent purity, we use the formula:
\(\text{Percent Purity} = \frac{\text{mass of pure substance}}{\text{total mass of sample}} \times 100\).
In the exercise, after calculating the mass of pure \(\text{NaOH}\) from the given sample, we use this formula to determine the purity percentage of \(\text{NaOH}\) in the sample, which we found to be approximately 87.82%.
\(\text{Percent Purity} = \frac{\text{mass of pure substance}}{\text{total mass of sample}} \times 100\).
In the exercise, after calculating the mass of pure \(\text{NaOH}\) from the given sample, we use this formula to determine the purity percentage of \(\text{NaOH}\) in the sample, which we found to be approximately 87.82%.
Mole Ratio
The mole ratio is a crucial concept in stoichiometry and it comes from the coefficients of a balanced chemical equation. It tells us the proportion of reactants and products involved in the reaction.
From the equation \(\text{NaOH} + \text{HCl} \rightarrow \text{NaCl} + \text{H}_2\text{O}\), the mole ratio of \(\text{NaOH}\) to \(\text{HCl}\) is 1:1. This means that one mole of \(\text{NaOH}\) reacts with one mole of \(\text{HCl}\).
Such ratios are essential for converting between moles of different substances in stoichiometric calculations.
From the equation \(\text{NaOH} + \text{HCl} \rightarrow \text{NaCl} + \text{H}_2\text{O}\), the mole ratio of \(\text{NaOH}\) to \(\text{HCl}\) is 1:1. This means that one mole of \(\text{NaOH}\) reacts with one mole of \(\text{HCl}\).
Such ratios are essential for converting between moles of different substances in stoichiometric calculations.
Molarity
Molarity is a measure of concentration of a solution, defined as the number of moles of solute per liter of solution. The formula for molarity is: \(\text{M} = \frac{\text{moles}}{\text{volume (L)}}\).
We used this formula in the exercise to find the moles of \(\text{HCl}\) using its given molarity and volume, which was necessary for further calculations in solving the problem. Understanding molarity is essential in preparing solutions and performing quantitative chemical analysis.
We used this formula in the exercise to find the moles of \(\text{HCl}\) using its given molarity and volume, which was necessary for further calculations in solving the problem. Understanding molarity is essential in preparing solutions and performing quantitative chemical analysis.
