Question

A 48.90-mL sample of a \(0.2040 \mathrm{M}\) acid reacts with an excess of \(\mathrm{Na}_{2} \mathrm{CO}_{3}\) to form \(125.0 \mathrm{~mL} \mathrm{CO}_{2}\) at \(722 \mathrm{mmHg}\) and \(17^{\circ} \mathrm{C}\). If the acid is either \(\mathrm{HCl}\) or \(\mathrm{H}_{2} \mathrm{SO}_{4},\) which is it?

Short answer

The acid is \(\mathrm{HCl}\).

Step-by-step solution

Convert Volume to Liters

First, convert the volume of the acid from milliliters to liters. Since there are 1000 milliliters in a liter, the volume in liters is 48.90 mL divided by 1000.\[ \text{Volume of acid in liters} = \frac{48.90}{1000} = 0.04890 \text{ L} \]

Calculate Moles of Acid

Use the molarity and volume in liters to calculate the moles of acid. The molarity formula is:\[ \text{Moles of acid} = \text{Molarity} \times \text{Volume in liters} \]So,\[ \text{Moles of acid} = 0.2040 \text{ M} \times 0.04890 \text{ L} = 0.009976 \text{ moles} \]

Convert Pressure to Atmospheres

Convert the pressure from mmHg to atmospheres. Use the conversion factor where 1 atm = 760 mmHg:\[ \text{Pressure in atm} = \frac{722 \text{ mmHg}}{760 \text{ mmHg/atm}} = 0.950 \text{ atm} \]

Convert Temperature to Kelvin

Convert the temperature from degrees Celsius to Kelvin by adding 273.15:\[ T = 17^{\circ}C + 273.15 = 290.15 \text{ K} \]

Use Ideal Gas Law to Find Moles of CO2

Use the ideal gas law, \(PV = nRT\), to calculate the moles of \(\mathrm{CO}_2\). Rearrange to solve for \(n\):\[ n = \frac{PV}{RT} \]Where:- \( P = 0.950 \text{ atm} \)- \( V = 125.0 \text{ mL} \equiv 0.1250 \text{ L} \)- \( R = 0.0821 \text{ L atm K}^{-1} \text{ mol}^{-1} \)- \( T = 290.15 \text{ K} \)Substitute these values into the equation:\[ n = \frac{(0.950 \text{ atm})(0.1250 \text{ L})}{(0.0821)}(290.15 \text{ K}) = 0.00505 \text{ moles of } \mathrm{CO}_2 \]

Determine the Acid Based on Stoichiometry

Determine which acid was used based on the stoichiometry of the reaction between the acid and \(\mathrm{Na}_2\mathrm{CO}_3\). The reactions are:1. \(2 \mathrm{HCl} + \mathrm{Na}_2\mathrm{CO}_3 \rightarrow 2 \mathrm{NaCl} + \mathrm{H}_2\mathrm{O} + \mathrm{CO}_2\)2. \(\mathrm{H}_2\mathrm{SO}_4 + \mathrm{Na}_2\mathrm{CO}_3 \rightarrow \mathrm{Na}_2\mathrm{SO}_4 + \mathrm{H}_2\mathrm{O} + \mathrm{CO}_2\)1. For \(\mathrm{HCl}\): - 2 moles of \(\mathrm{HCl}\) produce 1 mole of \(\mathrm{CO}_2\) \[ \mathrm{Moles \; of \; HCl} = 2 \times \mathrm{Moles \; of \; CO}_2 = 2 \times 0.00505 = 0.0101 \text{ moles} \] This is close to the calculated 0.009976 moles of acid.2. For \(\mathrm{H}_2\mathrm{SO}_4\): - 1 mole of \(\mathrm{H}_2\mathrm{SO}_4\) produces 1 mole of \(\mathrm{CO}_2\) \[ \mathrm{Moles \; of } \mathrm{H}_2\mathrm{SO}_4 = \mathrm{Moles \; of \; CO}_2 = 0.00505 \text{ moles} \] This is much lower than the calculated 0.009976 moles of acid.

Conclusion

The moles of acid required based on \(\mathrm{CO}_2\) production is closer to the moles of \(\mathrm{HCl}\) calculated than \(\mathrm{H}_2\mathrm{SO}_4\). Therefore, the acid is \(\mathrm{HCl}\).

Key concepts

Ideal Gas Law

The Ideal Gas Law is a fundamental equation in chemistry that describes the relationship between the pressure, volume, temperature, and amount of a gas. It's given by the formula: \[ PV = nRT \] where:

• \( P \) is the pressure of the gas in atmospheres.
• \( V \) is the volume of the gas in liters.
• \( n \) is the number of moles of gas.
• \( R \) is the ideal gas constant, approximately 0.0821 L atm K^{-1} mol^{-1}.
• \( T \) is the temperature of the gas in Kelvin.

In the context of the given problem, the Ideal Gas Law helps us find the moles of \(\mathrm{CO}_2\) produced in the reaction. This will allow us to determine which acid was used based on the stoichiometry of the reaction. By rearranging the Ideal Gas Law formula, we can solve for the moles of gas \( n \) using the experimental values of pressure, volume, and temperature. This is crucial for linking the measured volume of \( \mathrm{CO}_2 \) to the moles of acid that reacted.

Molarity

Molarity is a measure of concentration in chemistry, representing the number of moles of solute per liter of solution. It is given by the formula:\[ \text{Molarity} (M) = \frac{\text{moles of solute}}{\text{volume of solution in liters}} \]In the problem at hand, molarity allows us to calculate the moles of acid present in the given volume of solution. We start by converting the volume from milliliters to liters, and then use the molarity value to find the moles of acid.Molarity is a central concept in the problem because knowing the moles of acid involved in the reaction is necessary to compare with the moles of \(\mathrm{CO}_2\) produced. This comparison helps decide whether the acid in question is \(\mathrm{HCl}\) or \(\mathrm{H}_2\mathrm{SO}_4\). Understanding molarity ensures precise measurements for reacting substances in chemical equations, thereby helping predict the outcomes accurately.

Chemical Reaction

A chemical reaction involves the transformation of substances through breaking and forming chemical bonds, resulting in the production of new substances. In the given exercise, we explore two possible reactions involving acids and sodium carbonate. These are described as:

• \( 2 \mathrm{HCl} + \mathrm{Na}_2\mathrm{CO}_3 \rightarrow 2 \mathrm{NaCl} + \mathrm{H}_2\mathrm{O} + \mathrm{CO}_2 \)
• \( \mathrm{H}_2\mathrm{SO}_4 + \mathrm{Na}_2\mathrm{CO}_3 \rightarrow \mathrm{Na}_2\mathrm{SO}_4 + \mathrm{H}_2\mathrm{O} + \mathrm{CO}_2 \)

In each reaction, the acid reacts with sodium carbonate to produce carbon dioxide, water, and a salt. The chemical reaction provides insight into the stoichiometry, which is the quantitative relationship between reactants and products.In stoichiometry, balancing the chemical equation is vital to ensure that the number of atoms for each element is the same on both sides of the reaction. This balance also helps in estimating the amount of reactants needed and predicting the amount of products formed, including \(\mathrm{CO}_2\). Without understanding the concept of a chemical reaction, determining the nature of the acid used would be challenging.

Acid-Base Reaction

An acid-base reaction is a special kind of chemical reaction where an acid reacts with a base, often resulting in the formation of salt and water. In the exercise, we focus on the reaction of an acid with a carbonate base (\(\mathrm{Na}_2\mathrm{CO}_3\)).Key points about acid-base reactions:

• Acids provide hydrogen ions (H\(^+\)).
• Bases often provide hydroxide ions (OH\(^-\)), or react by accepting hydrogen ions.
• Carbonates like \(\mathrm{Na}_2\mathrm{CO}_3\) release \(\mathrm{CO}_2\) gas when they react with acids.

In this problem, the acid (either \(\mathrm{HCl}\) or \(\mathrm{H}_2\mathrm{SO}_4\)) reacts with the base \(\mathrm{Na}_2\mathrm{CO}_3\) to form \(\mathrm{CO}_2\), a common product in such reactions. Understanding this concept helps us predict the products of the reaction and is crucial in determining the unknown acid in the exercise.