Question

The molar mass of a certain metal carbonate, \(\mathrm{MCO}_{3}\) can be determined by adding an excess of \(\mathrm{HCl}\) acid to react with all the carbonate and then "back-titrating" the remaining acid with \(\mathrm{NaOH}\). (a) Write an equation for these reactions. (b) In a certain experiment, \(20.00 \mathrm{~mL}\) of \(0.0800 \mathrm{M} \mathrm{HCl}\) was added to a \(0.1022-\mathrm{g}\) sample of \(\mathrm{MCO}_{3}\). The excess \(\mathrm{HCl}\) required \(5.64 \mathrm{~mL}\) of \(0.1000 \mathrm{M}\) \(\mathrm{NaOH}\) for neutralization. Calculate the molar mass of the carbonate and identify \(\mathrm{M}\).

Short answer

The metal is Barium (Ba).

Step-by-step solution

Write the Chemical Reactions

The initial reaction involves the metal carbonate with hydrochloric acid: \[ \mathrm{MCO}_{3} + 2\mathrm{HCl} \rightarrow \mathrm{MCl}_{2} + \mathrm{H}_{2}O + \mathrm{CO}_{2} \]The excess \( \mathrm{HCl} \) is then neutralized by \( \mathrm{NaOH} \):\[ \mathrm{HCl} + \mathrm{NaOH} \rightarrow \mathrm{NaCl} + \mathrm{H}_{2}O \]

Calculate Moles of Original HCl

To find the moles of HCl originally added, use the formula:\[ \text{moles of HCl} = M \times V \]Where \( M \) is the concentration and \( V \) is the volume in liters. Thus:\[ \text{moles of HCl added} = 0.0800 \text{ M} \times 0.02000 \text{ L} = 0.0016 \text{ moles} \]

Calculate Moles of HCl Neutralized by NaOH

For this calculation, use the concentration and volume of NaOH:\[ \text{moles of NaOH} = 0.1000 \text{ M} \times 0.00564 \text{ L} = 0.000564 \text{ moles} \]Since \( \mathrm{HCl} \) and \( \mathrm{NaOH} \) react in a 1:1 ratio, the moles of HCl neutralized by NaOH is also 0.000564 moles.

Calculate Moles of HCl Reacted with MCO3

Subtract the moles of HCl neutralized by NaOH from the moles of original HCl:\[ \text{moles of HCl reacted with } \mathrm{MCO}_{3} = 0.0016 - 0.000564 = 0.001036 \text{ moles} \]Since each \( MCO_3 \) reacts with 2 moles of HCl, the moles of \( MCO_3 \) is:\[ \frac{0.001036}{2} = 0.000518 \text{ moles} \]

Calculate Molar Mass of MCO3

Use the mass of \( MCO_3 \) and its moles to find the molar mass:\[ \text{Molar Mass of } \mathrm{MCO}_{3} = \frac{0.1022 \text{ g}}{0.000518 \text{ moles}} \approx 197.49 \text{ g/mol} \]

Identify the Metal M

The molar mass of \( \mathrm{MCO}_{3} \approx 197.49 \text{ g/mol} \). Subtract the molar mass of \( \mathrm{CO}_{3} \), which is 60.01 g/mol:\[ M = 197.49 - 60.01 = 137.48 \text{ g/mol} \]Made to correspond to Barium (Ba), with an atomic mass close to 137.33 g/mol.

Key concepts

Chemical Reactions

Chemical reactions involve the transformation of reactants into products under the influence of various conditions, such as temperature, pressure, or the presence of catalysts. These reactions are evidenced by observable changes such as gas production, color change, temperature variation, or precipitate formation. In titration exercises, chemical reactions help determine the quantities of unknown substances.

In our given problem, two distinct chemical reactions occur. The first reaction involves the metal carbonate, denoted as \( \mathrm{MCO}_{3} \), reacting with hydrochloric acid (\( \mathrm{HCl} \)). This results in the formation of metal chloride (\( \mathrm{MCl}_{2} \)), water, and carbon dioxide. The equation for this reaction is:

\[ \mathrm{MCO}_{3} + 2\mathrm{HCl} \rightarrow \mathrm{MCl}_{2} + \mathrm{H}_{2}O + \mathrm{CO}_{2} \]

The second reaction involves the neutralization of any excess \( \mathrm{HCl} \) by sodium hydroxide (\( \mathrm{NaOH} \)) during the back-titration process. This is represented by the equation:

\[ \mathrm{HCl} + \mathrm{NaOH} \rightarrow \mathrm{NaCl} + \mathrm{H}_{2}O \]These reactions together help calculate the precise amount of metal carbonate in the sample.

Molar Mass Calculation

Molar mass calculation is critical in finding how much mass one mole of a chemical substance has. It is measured in grams per mole (g/mol) and involves adding up the atomic masses of each element present in the compound, based on the periodic table.

In our context of titration, the molar mass of the unknown substance, \( \mathrm{MCO}_{3} \), is determined by considering both the moles of the substance that reacted and its mass. Let's break this down:

• The moles of \( \mathrm{MCO}_{3} \) are calculated using the provided data on how it reacts with \( \mathrm{HCl} \).
• The weight of \( \mathrm{MCO}_{3} \) is given from the experiment's sample weight.

Ultimately, the calculation is given by:

\[ \text{Molar Mass of } \mathrm{MCO}_{3} = \frac{\text{mass of the sample}}{\text{moles of } \mathrm{MCO}_{3}}\]

In our case, the calculated molar mass helps us identify the unknown metal in the metal carbonate.

Stoichiometry

Stoichiometry deals with the quantitative relationships between the numbers of moles of reactants and products in a chemical reaction. It allows us to predict the amounts of substances consumed and produced in a given reaction. Understanding stoichiometry is fundamental when performing titration analyses to find the concentration or identity of an analyte.

In this titration problem, stoichiometry is used first to determine how much \( \mathrm{HCl} \) reacts with \( \mathrm{NaOH} \) and \( \mathrm{MCO}_{3} \). The reaction equation dictates that each mole of \( \mathrm{MCO}_{3} \) reacts with two moles of \( \mathrm{HCl} \), so:

\[ \mathrm{MCO}_{3} + 2\mathrm{HCl} \rightarrow \mathrm{Products} \]

Applying stoichiometry, we subtract the moles of \( \mathrm{HCl} \) that reacted with \( \mathrm{NaOH} \) from the initial moles of \( \mathrm{HCl} \) to find how much \( \mathrm{HCl} \) reacted with \( \mathrm{MCO}_{3} \). This then allows us to find the moles and eventually the molar mass of \( \mathrm{MCO}_{3} \). Understanding these relationships ensures that the calculations and conclusions drawn in the experiment are accurate and reliable.

Neutralization Reaction

Neutralization reactions involve an acid and a base reacting to form water and a salt. They are crucial in titration, particularly in back-titration techniques, as they help determine the concentration of an analyte when direct titration might be challenging.

In this specific experiment, a back-titration was used. Initially, a known excess of \( \mathrm{HCl} \) was added to react with \( \mathrm{MCO}_{3} \). Then, the remaining \( \mathrm{HCl} \) was neutralized by titrating with \( \mathrm{NaOH} \). The chemical equation representing this reaction is:

\[ \mathrm{HCl} + \mathrm{NaOH} \rightarrow \mathrm{NaCl} + \mathrm{H}_{2}O \]

Since \( \mathrm{HCl} \) and \( \mathrm{NaOH} \) react in a 1:1 molar ratio, the moles of \( \mathrm{NaOH} \) used in neutralization directly indicate the moles of \( \mathrm{HCl} \) leftover. This information is pivotal in calculating the moles of \( \mathrm{HCl} \) initially needed to fully react with the metal carbonate, thus clarifying the amount of \( \mathrm{MCO}_{3} \) that was present.