Question

An 8.65-g sample of an unknown group \(2 \mathrm{~A}\) metal hydroxide is dissolved in \(85.0 \mathrm{~mL}\) 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 \(56.9 \mathrm{~mL}\) 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 metal cation: \(\mathrm{Ca}^{2+}, \mathrm{Sr}^{2+}, \mathrm{Ba}^{2+}\) ?

Short answer

The molar mass of the unknown metal hydroxide is 121.63 g/mol, and the identity of the metal cation is Sr²⁺.

Step-by-step solution

Calculate moles of HCl used in titration

We know that 56.9 mL of the 2.50 M HCl solution has been used in the titration. First, we need to convert the volume of the HCl solution from mL to L: \[V_\text{HCl} = 56.9\, mL \times \frac{1\, L}{1000\, mL} = 0.0569\, L\] Now we can use the molarity of the HCl solution to find the moles of HCl: \[n_\text{HCl} = M_\text{HCl} \times V_\text{HCl} = 2.50\, M \times 0.0569\, L = 0.14225\, mol\]

Calculate moles of metal hydroxide

Since the unknown metal is from group 2A, the hydroxide will have a general formula and reaction with HCl as follows: \[\text{M}(\text{OH})_2(aq) + 2\, \text{HCl}(aq) \rightarrow \text{MCl}_2(aq) + 2\, H_2O(l)\] As the stoichiometry between the metal hydroxide and HCl is 1:2, the moles of metal hydroxide, \(n_\text{M(OH)_2}\), can be found by dividing the moles of HCl by 2: \[n_\text{M(OH)_2} = \frac{n_\text{HCl}}{2} = \frac{0.14225\, mol}{2} = 0.071125\, mol\]

Calculate the molar mass of the metal hydroxide

We're given the mass of the metal hydroxide to be 8.65 g. Now, we'll divide the mass of the metal hydroxide by the moles we calculated in step 2 to find the molar mass: \[M_\text{M(OH)_2} = \frac{8.65\, g}{0.071125\, mol} = 121.63\, \frac{g}{mol}\]

Identify the metal cation

We can now use the molar mass of the metal hydroxide to identify the associated metal cation. Comparing the molar mass to the molar masses of the group 2A hydroxides with Ca, Sr, and Ba, we find: \[\text{Ca}(\text{OH})_2: 40.08\, g/mol (\text{for Ca}) + 2 \times 17.01\, g/mol (\text{for OH}) = 74.10\, g/mol\] \[\text{Sr}(\text{OH})_2: 87.62\, g/mol (\text{for Sr}) + 2 \times 17.01\, g/mol (\text{for OH}) = 121.64\, g/mol\] \[\text{Ba}(\text{OH})_2: 137.33\, g/mol (\text{for Ba}) + 2 \times 17.01\, g/mol (\text{for OH}) = 171.35\, g/mol\] The molar mass of the metal hydroxide we calculated in step 3, 121.63 \(g/mol\), is closest to the molar mass of Sr(OH)2, 121.64 \(g/mol\). Therefore, the identity of the metal cation is Sr²⁺.

Key concepts

Molar Mass Calculation

When you have a chemical compound and you know its mass and the number of moles, you can find its molar mass. This is an essential step in various chemistry problems, including titration experiments. Molar mass (\(M\)) is calculated using the formula:\[M = \frac{\text{mass}}{\text{moles}}\]In our example, an unknown metal hydroxide had a given mass of 8.65 g. We determined the number of moles during the titration process. By substituting these values into the formula, we found the molar mass to be 121.63 g/mol. This calculation helps identify the substance by comparing it to known theoretical molar masses of similar compounds. Accuracy in determining the mass and moles is crucial here.

Metal Cation Identification

Identifying an unknown metal cation involves comparing the calculated molar mass of the metal hydroxide against known values for possible candidates. In the exercise, we considered possible group 2A metal cations, such as

• Calcium (Ca²⁺)
• Strontium (Sr²⁺)
• Barium (Ba²⁺)

Knowing the molar mass of the unknown compound, we compared it with

• Ca(OH)₂ = 74.10 g/mol
• Sr(OH)₂ = 121.64 g/mol
• Ba(OH)₂ = 171.35 g/mol

Our experimental molar mass was 121.63 g/mol, very close to Sr(OH)₂, which made the metal cation Strontium (Sr²⁺). This showcases the importance of stoichiometric precision in lab experiments, directly influencing the accuracy of chemical analysis.

Acid-Base Reaction

In titration, an acid-base reaction occurs where an acid reacts with a base. This reaction reaches the equivalence point when the number of moles of acid equals the number of moles of base. During the reaction, an acid-base indicator signals this point by changing color. In the exercise, hydrochloric acid (HCl) titrated the unknown metal hydroxide, which is a base, according to the balanced chemical equation:\[M(OH)_2(aq) + 2 HCl(aq) \rightarrow MCl_2(aq) + 2 H_2O(l)\]Here, each mole of the metal hydroxide reacts with two moles of hydrochloric acid, showcasing a classic acid-base titration involving a strong acid and a strong base. Understanding this process is crucial for calculating correct chemical quantities for further analysis.

Stoichiometry in Titration

Stoichiometry is key in understanding the quantifiable relationships in chemical reactions. In titrations, it's specifically used to determine the amount of unknown reactant. For our problem, stoichiometry helps us connect the moles of the titrant (HCl) used, with the moles of the unknown compound \(M(OH)_2\). The balanced chemical equation shows a 1:2 ratio, indicating that one mole of metal hydroxide reacts with two moles of HCl. This stoichiometric relationship allowed us to figure out the moles of metal hydroxide by dividing our moles of HCl by two:\[n_{\text{M(OH)}_2} = \frac{n_{\text{HCl}}}{2}\]Grasping stoichiometric concepts is essential not only in theory but also in routine lab practices, ensuring precise calculation and reaction usage.