Problem 88 - Chemistry: Atoms First

Question

What volume of \(0.08892 M\) HNO \(_{3}\) is required to react completely with 0.2352 g of potassium hydrogen phosphate? \(2 \mathrm{HNO}_{3}(a q)+\mathrm{K}_{2} \mathrm{HPO}_{4}(a q) \rightarrow \mathrm{H}_{2} \mathrm{PO}_{4}(a q)+2 \mathrm{KNO}_{3}(a q)\)

Step-by-Step Solution

Answer

First determine the moles of K2HPO4 (\(\frac{0.2352\, g}{174.18\, g/mol}\)), then use stoichiometry (\(2\) moles HNO3 per mole K2HPO4) to find moles of HNO3 required, and finally use the molarity of HNO3 to find the volume needed.

Step 1: Write down the balanced chemical equation

It's important to start with the balanced chemical equation for the reaction. The equation provided is already balanced: \[2 \mathrm{HNO}_{3}(aq) + \mathrm{K}_{2}\mathrm{HPO}_{4}(aq) \rightarrow \mathrm{H}_{2}\mathrm{PO}_{4}^{-}(aq) + 2 \mathrm{KNO}_{3}(aq)\].

Step 2: Calculate moles of potassium hydrogen phosphate

To find out how many moles of potassium hydrogen phosphate (K2HPO4) are present, use its molar mass and the mass given in the problem. The molar mass of K2HPO4 is \(174.18 g/mol\). Using the formula \[\text{moles} = \frac{\text{mass}}{\text{molar mass}}\], calculate the moles of K2HPO4: \[\text{moles K2HPO4} = \frac{0.2352\, g}{174.18\, g/mol}\].

Step 3: Use the stoichiometry to find moles of HNO3 needed

According to the balanced chemical equation, 2 moles of HNO3 react with 1 mole of K2HPO4. Set up a proportion to find the number of moles of HNO3 needed.\[\text{moles HNO3 required} = 2 \times \text{moles K2HPO4}\].

Step 4: Calculate the volume of HNO3 solution required

Use the concentration of the HNO3 solution to find the volume required to provide the moles of HNO3 you calculated. Using the formula \[\text{volume} = \frac{\text{moles}}{\text{Molarity}}\], calculate the volume of HNO3: \[\text{volume HNO3 required} = \frac{\text{moles HNO3 required}}{0.08892\, M}\].

Step 5: Combine the calculations to find the final answer

Combine the calculations from Steps 2, 3, and 4 to determine the final volume of HNO3 required. First find the moles of K2HPO4, then use the stoichiometry to find moles of HNO3, and finally divide by the molarity of HNO3 to find the volume.

Key Concepts

Chemical Reaction Calculations

At the heart of chemistry lie chemical reaction calculations, which are vital for understanding how substances interact and transform. These calculations allow us to predict the amounts of reactants needed to produce a given quantity of product or to determine the quantity of product formed from certain reactants. For instance, in our exercise, we needed to determine what volume of nitric acid (HNO3) would react with a given mass of potassium hydrogen phosphate (K2HPO4).

To perform these calculations, a balanced chemical equation is essential as it provides the ratio of reactants to products, known as the stoichiometry of the reaction. This knowledge serves as a map, guiding us from the amount of one substance to the expected amount of another.

Molar Mass

Molar mass, the bridge between the microscopic world of atoms and the macroscopic world we can measure, is the weight of one mole of a substance, typically expressed in grams per mole (g/mol). This seemingly simple concept is crucial in moving from the mass of a substance to the amount in moles, which is the key to all stoichiometric calculations.

In our exercise, the molar mass of potassium hydrogen phosphate (K2HPO4) allowed us to convert the given mass (0.2352 g) into moles. It's like knowing the price per pound to calculate how much several apples will cost; molar mass tells us the 'price' per mole.

Molarity

Molarity, marked by the symbol M, reflects the concentration of a solution, providing the number of moles of solute per liter of solution. It's an invaluable figure when preparing or working with chemical solutions because it helps us understand the strength of the solution in terms of the amount of substance it contains.

In our example, we worked with a 0.08892 M solution of HNO3. Knowing the molarity enabled us to find the volume of this solution needed to obtain the moles of HNO3 necessary for the reaction. Molarity is akin to a recipe, telling us how much of each ingredient we need to mix to get the desired taste (or in this case, reaction).

Balanced Chemical Equations

Balanced chemical equations are the bedrock of stoichiometry. An equation is balanced when the number of atoms of each element is the same on both sides of the reaction. This balance reflects the conservation of mass principle: atoms are neither created nor destroyed in a chemical reaction. They're simply rearranged.

A balanced chemical equation is like a well-structured musical score, where each note (atom) has its place, ensuring harmony (balance). It tells us how many parts of each reactant interact and how many parts of each product form. For our exercise, the balanced equation dictated the exact ratio of reactants to products, which is essential to calculate how much HNO3 was required for the given amount of K2HPO4.