Chapter 14: Problem 85
If \(150 \mathrm{~mL}\) of \(0.055 \mathrm{M} \mathrm{HNO}_{3}\) are needed to completely neutralize \(1.48 \mathrm{~g}\) of an impure sample of sodium hydrogen carbonate (baking soda), what percent of the sample is baking soda?
Short Answer
Expert verified
The percent of the sample that is baking soda is equal to the mass of pure \(NaHCO_3\) divided by the total mass of the sample, multiplied by 100.
Step by step solution
01
Write the balanced chemical equation
Start by writing the chemical reaction between nitric acid \(HNO_3\) and sodium hydrogen carbonate \(NaHCO_3\). The balanced equation is: \[ HNO_3 + NaHCO_3 \rightarrow NaNO_3 + CO_2 + H_2O \]
02
Calculate moles of \(HNO_3\) used
Since molarity \(M\) is moles of solute \(n\) per liter of solution \(V\), calculate the moles of nitric acid with the formula: \[ n_{HNO_3} = M_{HNO_3} \times V_{HNO_3} = 0.055 \text{ M} \times 0.150 \text{ L} \]
03
Determine moles of \(NaHCO_3\) based on stoichiometry
The balanced equation shows a 1:1 mole ratio between \(HNO_3\) and \(NaHCO_3\), hence the moles of \(NaHCO_3\) will be equal to the moles of \(HNO_3\).
04
Calculate the mass of pure \(NaHCO_3\) in grams
Use the molar mass of \(NaHCO_3\) (84.01 g/mol) to convert moles to grams with the formula: \[ m_{NaHCO_3} = n_{NaHCO_3} \times M_{NaHCO_3} = n_{HNO_3} \times 84.01 \text{ g/mol} \]
05
Calculate the percent purity of the sample
The percent purity is the mass of pure \(NaHCO_3\) divided by the mass of the impure sample, multiplied by 100 to get the percentage: \[ \text{Percent purity} = \left( \frac{m_{NaHCO_3}}{1.48 \text{ g}} \right) \times 100\% \]
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Balanced Chemical Equation
Understanding a balanced chemical equation is crucial in the study of chemistry, as it represents the law of conservation of mass. It illustrates that atoms are neither created nor destroyed in a chemical reaction. In our sample exercise, the balanced equation \( HNO_3 + NaHCO_3 \rightarrow NaNO_3 + CO_2 + H_2O \) ensures that the number of atoms for each element is the same on both sides of the equation.
When working with chemical reactions, always check the balance of your equation first. This will tell you the stoichiometric ratios needed to calculate reactants and products – essential knowledge for subsequent calculations in the problem.
When working with chemical reactions, always check the balance of your equation first. This will tell you the stoichiometric ratios needed to calculate reactants and products – essential knowledge for subsequent calculations in the problem.
Molarity Calculation
Molarity, symbolized as \(M\), measures the concentration of a solution and is defined as the number of moles of solute per liter of solution. It is a fundamental concept when dealing with chemical solutions, as it allows us to relate the volume of a solution to the amount of substance in moles. The formula \( n_{HNO_3} = M_{HNO_3} \times V_{HNO_3} \) demonstrates this calculation. Here, knowing the molarity and volume of \(HNO_3\) solution enables us to find the moles of \(HNO_3\) used in the reaction.
Keep in mind, the volume needs to be in liters for the calculation to work correctly. If provided in milliliters, as in our problem, you must convert to liters by dividing by 1000.
Keep in mind, the volume needs to be in liters for the calculation to work correctly. If provided in milliliters, as in our problem, you must convert to liters by dividing by 1000.
Mole-to-Mass Conversion
Once the moles of a substance are known, mole-to-mass conversion allows us to determine its mass, which is vital in quantitative chemistry. This step utilizes the substance's molar mass. The relationship is given by \( m = n \times M_{molar} \) where \(m\) is the mass, \(n\) is the number of moles, and \(M_{molar}\) is the molar mass.
In the given exercise, after finding the moles of \(NaHCO_3\), we multiply by its molar mass (84.01 g/mol) to calculate the mass. This step is essential in moving from the abstract quantity (moles) to a tangible quantity (grams) that we can measure.
In the given exercise, after finding the moles of \(NaHCO_3\), we multiply by its molar mass (84.01 g/mol) to calculate the mass. This step is essential in moving from the abstract quantity (moles) to a tangible quantity (grams) that we can measure.
Percent Purity Calculation
The percent purity of a chemical compound is a measure of its composition in terms of pure substance. It's particularly important in industry where the quality of materials matters. Calculated using the formula \( \text{Percent purity} = (\frac{m_{pure}}{m_{impure}}) \times 100\% \) where \(m_{pure}\) is the mass of the pure substance and \(m_{impure}\) is the mass of the impure sample.
In our problem, to find the purity of the baking soda sample, we calculate the mass of pure \(NaHCO_3\) and express it as a percentage of the initial mass of the sample provided in the problem. This tells us how much of the original sample is the desired compound.
In our problem, to find the purity of the baking soda sample, we calculate the mass of pure \(NaHCO_3\) and express it as a percentage of the initial mass of the sample provided in the problem. This tells us how much of the original sample is the desired compound.
Acid-Base Neutralization
The neutralization reaction between an acid and a base is a specific type of double displacement reaction, where an acid reacts with a base to produce a salt and water. This reaction invariably involves the transfer of protons from the acid to the base. In the context of our example, nitric acid \(HNO_3\) reacts with sodium hydrogen carbonate \(NaHCO_3\) following the equation shown above.
Understanding the stoichiometry of neutralization reactions is key for calculations, as it often involves a 1:1 mole ratio for monoprotic acids and bases, simplifying calculations. For multiprotic acids or bases, the ratio might differ, which is essential to consider when balancing equations.
Understanding the stoichiometry of neutralization reactions is key for calculations, as it often involves a 1:1 mole ratio for monoprotic acids and bases, simplifying calculations. For multiprotic acids or bases, the ratio might differ, which is essential to consider when balancing equations.