Problem 89 - General Chemistry

Question

What volume of \(0.250 \mathrm{M} \mathrm{HNO}_{3}\) (nitric acid) reacts with \(44.8 \mathrm{~mL}\) of \(0.150 \mathrm{M} \mathrm{Na}_{2} \mathrm{CO}_{3}\) (sodium carbonate) in the following reaction? $$ 2 \mathrm{HNO}_{3}(a q)+\mathrm{Na}_{2} \mathrm{CO}_{3}(a q) \longrightarrow{2 \mathrm{NaNO}_{3}(a q)+\mathrm{H}_{2} \mathrm{O}(l)+\mathrm{CO}_{2}(g)} $$

Step-by-Step Solution

Answer

53.76 mL of 0.250 M HNO3 is needed.

Step 1: Understand the Stoichiometry of the Reaction

The balanced equation is \( 2 \mathrm{HNO}_{3} + \mathrm{Na}_{2} \mathrm{CO}_{3} \rightarrow 2 \mathrm{NaNO}_{3} + \mathrm{H}_{2} \mathrm{O} + \mathrm{CO}_{2} \). This tells us that 2 moles of nitric acid \((\mathrm{HNO}_3)\) react with 1 mole of sodium carbonate \((\mathrm{Na}_2\mathrm{CO}_3)\).

Step 2: Calculate Moles of Sodium Carbonate

Use the concentration and volume of sodium carbonate to find moles: \( \text{moles of } \mathrm{Na}_2\mathrm{CO}_3 = 0.150 \, \mathrm{M} \times \frac{44.8}{1000} \, \mathrm{L} = 0.00672 \, \text{moles} \).

Step 3: Determine Required Moles of Nitric Acid

From the stoichiometry (2 mole \( \mathrm{HNO}_3 \) per mole of \( \mathrm{Na}_2\mathrm{CO}_3 \)), calculate moles of \( \mathrm{HNO}_3 \) needed: \( 0.00672 \, \text{moles of } \mathrm{Na}_2\mathrm{CO}_3 \times \frac{2}{1} = 0.01344 \, \text{moles of } \mathrm{HNO}_3 \).

Step 4: Calculate Volume of Nitric Acid Solution

Use the moles of nitric acid and its concentration to find the volume: \( \text{Volume} = \frac{0.01344 \, \text{moles}}{0.250 \, \mathrm{M}} = 0.05376 \, \mathrm{L} \), or \( 53.76 \, \mathrm{mL} \).

Key Concepts

Molarity

Molarity is an essential concept in chemistry referring to the concentration of a solute in a solution. It is defined as the number of moles of a solute divided by the volume of the solution in liters. Molarity is usually expressed in units of moles per liter (mol/L) and is often symbolized by 'M'.
Understanding molarity helps us know the strength of a solution, which is crucial in various chemical calculations like determining how much reactant is needed or how much product will be formed. For example, in the problem, the molarity of

• Nitric acid (HNO₃) is given as 0.250 M
• Sodium carbonate (Na₂CO₃) has a molarity of 0.150 M

Molarity is vital in stoichiometric calculations to convert between the volume of solutions and the amount of a chemical substance.

Chemical Reactions

Chemical reactions represent the process where reactants transform into products. They are characterized by the rearrangement of atoms and are usually represented through chemical equations.
In our given exercise, the chemical reaction between nitric acid (HNO₃) and sodium carbonate (Na₂CO₃) results in the formation of sodium nitrate (NaNO₃), water (H₂O), and carbon dioxide (CO₂).
Chemical reactions can be classified into different types, such as synthesis, decomposition, single replacement, and double replacement. The reaction here is a double-replacement reaction where ions from two reactants exchange to form new products.

Balanced Chemical Equations

Balanced chemical equations are vital as they accurately represent what happens during a chemical reaction, showing the conservation of mass. It ensures the number of atoms for each element is the same on both sides of the equation.
In the exercise, the balanced equation is:\[2 \text{HNO}_3 (aq) + \text{Na}_2\text{CO}_3 (aq) \rightarrow 2 \text{NaNO}_3 (aq) + \text{H}_2\text{O} (l) + \text{CO}_2 (g)\]Here, for example, there are 2 moles of nitric acid reacting with 1 mole of sodium carbonate. This balance is important for calculating how much of each reactant is needed and how much product is formed.
Balancing equations requires a systematic approach where you balance elements that appear in the least compounds first and adjust coefficients until the atoms for each element are equal on both sides.

Volume Calculation

Calculating volume is a common and practical application in chemistry. It involves using the relationship between moles, molarity, and volume to solve various chemical problems.
In the problem, we first calculated the moles of sodium carbonate using its molarity and volume:\[\text{Moles of } \text{Na}_2\text{CO}_3 = 0.150 \text{ M} \times \frac{44.8}{1000} \text{ L} = 0.00672 \text{ moles}\]Afterwards, using the balanced equation, the moles of nitric acid needed were found to be 0.01344 moles. Finally, with the molarity of nitric acid provided as 0.250 M, the volume needed was calculated as:\[\text{Volume} = \frac{0.01344 \text{ moles}}{0.250 \text{ M}} = 0.05376 \text{ L} = 53.76 \text{ mL}\]This process highlights the importance of understanding molarity and chemical stoichiometry for accurately determining solution volumes in reactions.