Problem 82 - Chemistry The Central Science

Question

(a) How many milliliters of 0.120 \(\mathrm{M}\) HCl are needed to completely neutralize 50.0 \(\mathrm{mL}\) of 0.101 \(\mathrm{M} \mathrm{Ba}(\mathrm{OH})_{2}\) solution? (b) How many milliliters of 0.125 \(\mathrm{M} \mathrm{H}_{2} \mathrm{SO}_{4}\) are needed to neutralize 0.200 \(\mathrm{g}\) of NaOH? \((\mathrm{c})\) If 55.8 \(\mathrm{mL}\) of a BaCl \(_{2}\) solution is needed to precipitate all the sulfate ion in a 752 -mg sample of \(\mathrm{Na}_{2} \mathrm{SO}_{4},\) what is the molarity of the BaCl\(_{2}\) solution?

Step-by-Step Solution

Answer

In a short answer, (a) 84.2 mL of 0.120 M HCl are needed to completely neutralize 50.0 mL of 0.101 M Ba(OH)$_2$ solution. (b) 20.0 mL of 0.125 M H$_2$SO$_4$ are needed to neutralize 0.200 g of NaOH. (c) The molarity of the BaCl$_2$ solution is 0.095 M.

Step 1: Individual tasks

1. Calculate the number of moles 2. Convert the moles to volume (where applicable) 3. Proceed to the next problem

Step 2: Problem a (part 1): Determine the Amount of Moles for HCl and \( \text{Ba(OH)}_2 \)

The balanced chemical equation will come out to be \[2\text{HCl} + \text{Ba(OH)}_2 \rightarrow \text{BaCl}_2 + 2\text{H}_2\text{O}\] In 50.0 mL of 0.101 M \(\text{Ba(OH)}_2\), the number of moles is: \(0.101 M \times 0.050 L = 0.00505 \text{ mole of Ba(OH)}_2\). Since the reaction with HCl is in a 2:1 ratio, the number of moles of HCl required will be \(2 \times 0.00505 = 0.0101 \text{ moles of HCl}\).

Step 3: Problem a (part 2): Convert Moles to Volume for HCl

To convert this into volume, you need to use the molarity definition, M = moles/L. With reorganization you get V = moles / M, substituting the values: \(0.0101 \text{ moles} / 0.120 \text{M} = 0.0842L = 84.2 \text{mL of HCl}\).

Step 4: Problem b (part 1): Determine Mole for \( \text{NaOH} \) and \( \text{H}_2\text{SO}_4 \)

The balanced equation for the neutralization of \( \text{NaOH} \) and \( \text{H}_2\text{SO}_4 \) is: \[2\text{NaOH} + \text{H}_2\text{SO}_4 \rightarrow \text{Na}_2\text{SO}_4 + 2\text{H}_2\text{O}\] 0.200 g of \( \text{NaOH} \), which equates to \( \text{NaOH: 0.200 g} / 40.0 g\text{/mol} = 0.00500 \text{ moles of NaOH} \). The stoichiometry from the balanced equation tells us that 2 moles of \( \text{NaOH} \) reacts with 1 mole of \( \text{H}_2\text{SO}_4 \). So, we will need \( 0.00500 \text{ moles of NaOH} / 2 = 0.00250 \text{ moles of H}_2\text{SO}_4 \).

Step 5: Problem b (part 2): Convert Moles to Volume for \( \text{H}_2\text{SO}_4 \)

The same as in part a, \( V = \text{moles} / \text{M} \). Substituting we find \( 0.00250 \text{ moles} / 0.125 \text{M} = 0.0200 \text{L} = 20.0 \text{mL of H}_2\text{SO}_4 \).

Step 6: Problem c (part 1): Determine moles of \( \text{Na}_2\text{SO}_4 \) and \( \text{BaCl}_2 \)

The balanced equation for the reaction between the sulfate ion from \( \text{Na}_2\text{SO}_4 \) and \( \text{BaCl}_2 \) is: \[\text{BaCl}_2 + \text{Na}_2\text{SO}_4 \rightarrow \text{BaSO}_4 + 2\text{NaCl}\] 752 mg of \( \text{Na}_2\text{SO}_4 \) would be equivalent to \( \text{Na}_2\text{SO}_4: 0.752 g / 142.04 g\text{/mol} = 0.00529 \text{ moles} \). The balanced equation suggests that 1 mole of \( \text{Na}_2\text{SO}_4 \) reacts with 1 mole of \( \text{BaCl}_2 \) to produce the precipitate. So, we will need the equivalent 0.00529 moles of \( \text{BaCl}_2 \).

Step 7: Problem c (part 2): Determine molarity of \( \text{BaCl}_2 \)

Molarity (M) is given by the formula \( M = \text{moles} / V \). Using the given volume and the calculated moles of \( \text{BaCl}_2 \), we find, \( M = 0.00529 \text{ moles} / 0.0558 \text{L} = 0.095 \text{M} \) for the \( \text{BaCl}_2 \) solution.

Key Concepts

Molarity Calculations

Molarity is a way to express the concentration of a solution. It is the number of moles of solute per liter of solution. The formula for calculating molarity is given by \( M = \frac{\text{moles of solute}}{\text{liters of solution}} \). This simple formula allows us to determine how concentrated a solution is.

For instance, if you have a 0.120 M hydrochloric acid (HCl) solution and you need a certain volume to neutralize a base, you can rearrange the formula to find the volume: \( V = \frac{\text{moles}}{M} \). This formula was applied in the exercise to find the milliliters of HCl needed. Remember, the key is knowing the number of moles involved in the reaction and the molarity of the solution you're working with.

Similarly, when converting from moles to volume or vice versa in a chemical reaction, molarity calculations become a foundational tool. An understanding of molarity helps you relate the chemical equation to real-life measurements.

Stoichiometry

Stoichiometry is the method used to calculate the quantities of reactants and products in a chemical reaction. It relies on the balanced chemical equations that represent reactions. These equations give the mole ratio between reactants and products, which is crucial for calculations.

For example, in the neutralization reaction between HCl and \( \text{Ba(OH)}_2 \), the balanced equation \( 2\text{HCl} + \text{Ba(OH)}_2 \rightarrow \text{BaCl}_2 + 2\text{H}_2\text{O} \) tells us that it takes 2 moles of HCl to neutralize 1 mole of \( \text{Ba(OH)}_2 \). This ratio is key whenever you adjust the amount of one substance to find the amount of another.

Stoichiometry involves calculating the moles of each compound using their respective molar masses and mole ratios. In Problem b, we interpreted the stoichiometry to determine the amount of \( \text{H}_2\text{SO}_4 \) needed to react with a known mass of \( \text{NaOH} \). By using unit conversions and the mole ratios from the balanced equation, students can accurately predict how chemicals will react.

Chemical Equations

Chemical equations are symbolic representations of chemical reactions. They show the reactants and products as well as their proportions in terms of moles. Balancing these equations is essential as it ensures the law of conservation of mass is upheld, meaning matter is neither created nor destroyed in the process.

Each equation comprises actual compounds along with their states (e.g., solid, liquid). For example, when \( \text{NaOH} \) reacts with \( \text{H}_2\text{SO}_4 \), the balanced equation \( 2\text{NaOH} + \text{H}_2\text{SO}_4 \rightarrow \text{Na}_2\text{SO}_4 + 2\text{H}_2\text{O} \) shows how sodium hydroxide and sulfuric acid produce sodium sulfate and water. This chemical equation helps to determine the stoichiometry and thereby the quantity of each substance required or produced.

In practice, writing and balancing chemical equations demand an understanding of the roles each atom and molecule play. It serves as a roadmap for understanding the reactants needed, the products formed, as well as their ratios and reaction conditions. Getting comfortable with chemical equations empowers students to explore more complex reactions and predict their outcomes accurately.