Problem 85 - Chemistry The Central Science In Si Units

Question

A \(4.36-g\) sample of an unknown alkali metal hydroxide is dissolved in \(100.0 \mathrm{~mL}\) of water. An acid-base indicator is added, and the resulting solution is titrated with \(2.50 \mathrm{M} \mathrm{HCl}(a q)\) solution. The indicator changes color, signaling that the equivalence point has been reached, after \(17.0 \mathrm{~mL}\) of the hydrochloric acid solution has been added. (a) What is the molar mass of the metal hydroxide? (b) What is the identity of the alkali metal cation: \(\mathrm{Li}^{+}, \mathrm{Na}^{+}, \mathrm{K}^{+}, \mathrm{Rb}^{+},\) or \(\mathrm{Cs}^{+} ?\)

Step-by-Step Solution

Answer

(a) Molar mass: 102.588 g/mol; (b) the alkali metal is Rubidium (Rb).

Step 1: Determine Moles of HCl Used

Calculate the moles of HCl that were used to reach the equivalence point using the formula \( moles = Molarity \times Volume \). Convert the volume from milliliters to liters: \( 17.0 \text{ mL} = 0.0170 \text{ L} \). Therefore,\[moles\ ext{HCl} = 2.50 \text{ M} \times 0.0170 \text{ L} = 0.0425 \text{ moles}.\]

Step 2: Relate Moles of HCl to Moles of Metal Hydroxide

Since the reaction between an alkali metal hydroxide \(( ext{MOH} )\) and HCl is a 1:1 stoichiometric reaction (\( ext{MOH} + ext{HCl} ightarrow ext{MCl} + ext{H}_2 ext{O} \)), the moles of \( ext{MOH}\) are equal to the moles of HCl. Thus, the moles of \( ext{MOH}\) are 0.0425.

Step 3: Calculate the Molar Mass of Metal Hydroxide

The molar mass of the metal hydroxide \( ext{MOH} \) can be determined by dividing the sample mass by the number of moles:\[\text{Molar mass of} \ ext{MOH} = \frac{4.36 \text{ g}}{0.0425 \text{ moles}} = 102.588 \text{ g/mol}.\]

Step 4: Identify the Alkali Metal Cation

Knowing the molar mass of \( ext{MOH} = 102.588 \text{ g/mol} \), calculate the mass of \( ext{OH}^- \) group, which is 17.01 \text{ g/mol}. Subtract this from the \( ext{MOH} \) molar mass to find the atomic mass of the metal \( ext{M} \):\[\text{Molar mass of} \ ext{M} = 102.588 - 17.01 = 85.578 \text{ g/mol}.\]This atomic mass approximately corresponds to the molar mass of Rb \(( ext{Rubidium} )\).

Step 5: Conclusion

The molar mass suggests that the metal is rubidium, as the calculated mass of \(85.578 \text{ g/mol}\) is closest to rubidium's standard atomic weight, which is around 85.47 \text{ g/mol}.

Key Concepts

Alkali Metals

Alkali metals are a group of elements found in Group 1 of the periodic table. These elements include lithium (Li), sodium (Na), potassium (K), rubidium (Rb), and cesium (Cs). They are known for being highly reactive, especially with water, and can form strong bases when combined with hydroxide ions (OH).

Alkali metals have one electron in their outermost shell, which makes them eager to donate that electron in chemical reactions. This characteristic accounts for their high reactivity.

• They form hydroxides that are very soluble in water, creating basic solutions with high pH values.
• Each alkali metal has a unique atom size and atomic mass, with increasing atomic mass down the group.
• Rubidium, for instance, closely aligns with the calculated molar mass in our exercise, suggesting its presence in the sample.

Molar Mass Calculation

Calculating the molar mass involves determining the mass of one mole of a compound. This is essential for understanding stoichiometry in chemical reactions.

In the provided exercise, the molar mass of an unknown alkali metal hydroxide was found using the mass of the sample and the moles derived from the titration:

• First, determine the moles of the hydroxide using the stoichiometry of the reaction. Here, the reaction with HCl was a 1:1 ratio.
• Next, molar mass is calculated by dividing the given mass of the hydroxide sample by the moles of hydroxide.
• This calculation helps identify the unknown metal by comparing calculated values with known atomic masses of alkali metals.

Acid-Base Reaction

Acid-base reactions are fundamental chemical reactions where an acid and a base interact to form water and a salt. In the titration experiment described, an alkali metal hydroxide acted as the base, (\[ ext{MOH (s)} + ext{HCl (aq)} \rightarrow ext{MCl (aq)} + ext{H}_2 ext{O (l)}\]), engaging in a neutralization reaction with hydrochloric acid.

• Titrations are practical applications that help determine the concentration of an unknown solution by reacting it with a solution of known concentration.
• The moment enough acid has been added to exactly react with all the base, reaching the equivalence point, this reaction stops. The presence of an indicator helps signal this completion by a color change.

Equivalence Point

The equivalence point in a titration is crucial as it signifies the moment when the amount of added titrant is stoichiometrically equivalent to the amount of analyte in the sample. In the case of acid-base titrations, it is the point at which the amount of acid equals the amount of base in the solution.

• Reaching the equivalence point indicates that the reaction between the acid and base is complete.
• An indicator is typically used to visually signal this point; a color change occurs when this balance is achieved.
• In the exercise, the equivalence point was reached after adding 17.0 mL of 2.50 M HCl.

Knowing the equivalence point allows chemists to accurately calculate unknown concentrations and determine characteristics about the substances involved in the reaction.