Problem 82 - General Chemistry: Principles and Modern Applications

Question

The pH of saturated \(\operatorname{Sr}(\text { OH })_{2}(\text { aq })\) is found to be 13.12 A \(10.0 \mathrm{mL}\) sample of saturated \(\operatorname{Sr}(\mathrm{OH})_{2}(\mathrm{aq})\) is diluted to \(250.0 \mathrm{mL}\) in a volumetric flask. A \(10.0 \mathrm{mL}\) sample of the diluted \(\operatorname{Sr}(\mathrm{OH})_{2}(\mathrm{aq})\) is transferred to a beaker, and some water is added. The resulting solution requires \(25.1 \mathrm{mL}\) of a \(\mathrm{HCl}\) solution for its titration. What is the molarity of this HCl solution?

Step-by-Step Solution

Answer

The molarity of the hydrochloric acid solution is 0.021 mol/L.

Step 1: Calculate the concentration of \(\operatorname{Sr}(\mathrm{OH})_{2}\)

Since pH of saturated \(\operatorname{Sr}\left(\mathrm{OH}\right)_{2}\) is given as 13.12, we can calculate concentration of \(\operatorname{Sr}(\mathrm{OH})_{2}\) using formula for pOH and pH. The pOH is found by subtracting given pH from 14 (since pH + pOH = 14), which gives \(pOH = 14 - 13.12 = 0.88 \). The \(\operatorname{OH}^{-}\) concentration (molarity) is found by taking anti-log of \(-pOH\), so concentration of \(\operatorname{OH}^{-} = 10^{-0.88} \approx 0.132 \, mol \, L^{-1}\). As the formula of \(\operatorname{Sr}(\mathrm{OH})_{2}\) indicates, one molecule of it contains two \(\operatorname{OH}^{-}\) ions, the concentration of \(\operatorname{Sr}(\mathrm{OH})_{2}\) is half of the \(\operatorname{OH}^{-}\) molarity, which is \( 0.132/2 = 0.066 \, mol \, L^{-1}\).

Step 2: Determine new concentration after dilution

The initial volume of the \(\operatorname{Sr}\left(\mathrm{OH}\right)_{2}\) solution is 10.0 mL and it is diluted to 250.0 mL. We can use the dilution formula: \(M1 \times V1 = M2 \times V2 \), where \(M1\) is initial molarity (0.066 mol/L), \(V1\) is initial volume (10.0 mL), \(V2\) is final volume (250.0 mL), and \(M2\) is the final molarity we need to find out. So, \(M2 = (M1 \times V1) / V2 = (0.066 \times 10.0) / 250.0 = 0.00264 \, mol \, L^{-1}\). This is the molarity of \(\operatorname{Sr}\left(\mathrm{OH}\right)_{2}\) in the final 250 mL solution.

Step 3: Calculate the molarity of HCl

In the titration, 10.0 mL of the diluted \(\operatorname{Sr}\left(\mathrm{OH}\right)_{2}\) solution was used, containing \(\operatorname{Sr}\left(\mathrm{OH}\right)_{2}\) of concentration \(0.00264 \, mol \, L^{-1}\). The stoichiometry of the reaction between \(\operatorname{Sr}\left(\mathrm{OH}\right)_{2}\) and \(\mathrm{HCl}\) is 1:2. Which is why, twice the moles of \(\mathrm{HCl}\) will be used for the titration. So, molarity of \(\mathrm{HCl}\) (let's denote it as \(M\)) is given by \((2 \times moles \, of \, \operatorname{Sr}\left(\mathrm{OH}\right)_{2}) / volume \, of \, HCl \, used\) which equals \((2 \times 0.00264 \, mol \, L^{-1} \times 0.01 \, L) / 0.0251 \, L = 0.021 \, mol \, L^{-1}\). This is the molarity of the \(\mathrm{HCl}\) solution.

Key Concepts

Molarity Calculation

Molarity is a key concept in chemistry that helps us understand the concentration of a solution. It is defined as the number of moles of solute per liter of solution. This allows chemists to express concentrations precisely and predict how substances will react in solution.
To calculate molarity, use the formula:

• Molarity (\( M\) ) = moles of solute / liters of solution

In the original exercise, the concentration of strontium hydroxide, \( \operatorname{Sr}(\mathrm{OH})_{2}\), was determined indirectly using the pH value provided. Given the pH, we calculated the pOH by subtracting the pH from 14, leading to the \( \text{OH}^{-} \) ion concentration using the formula \( 10^{-\text{pOH}} \). Because each molecule of \( \operatorname{Sr}(\mathrm{OH})_{2}\) provides two \( \text{OH}^{-} \) ions, the concentration of the \( \operatorname{Sr}(\mathrm{OH})_{2}\) is half the concentration of the hydroxide ions.
This approach allows us to adapt and find the appropriate molar amounts needed for further calculations in reactions and dilutions.

pH and pOH Relationship

The relationship between pH and pOH is fundamental in understanding acidic and basic solutions. represents the acidity of a solution, while reflects its basicity. Together, they are related by the equation:

• \[ \text{pH} + \text{pOH} = 14 \]

This equation applies to all aqueous solutions at 25°C, making it a very handy tool in chemistry.
The pH value can be determined using the concentration of hydrogen ions, \( \text{H}^{+} \), through the equation \( \text{pH} = -\log[\text{H}^{+}] \). Similarly, the pOH is found using the hydroxide ions, \( \text{OH}^{-} \), through \( \text{pOH} = -\log[\text{OH}^{-}] \). By knowing either the pH or pOH, the other can easily be calculated.
In situations where the pH is known, finding the pOH helps to complete the picture of the solution’s properties, as was done in the original exercise with \( \operatorname{Sr}(\mathrm{OH})_{2} \)'s concentration calculation.

Dilution Formula

Dilution is an essential concept for preparing solutions of desired concentrations or altering the concentration of an existing solution. It's useful in laboratory settings, particularly when handling concentrated stock solutions.
One can calculate dilution using the formula:

• \[ M_1 \times V_1 = M_2 \times V_2 \]

Where \( M_1 \) and \( V_1 \) are the molarity and volume of the initial solution, and \( M_2 \) and \( V_2 \) are the desired molarity and volume after dilution.
In the exercise, we used this formula to determine how the concentration of \( \operatorname{Sr}(\mathrm{OH})_{2} \) changed after it was diluted from 10 mL to 250 mL. This calculation is crucial for ensuring that experimental results are accurate and calibrated properly.
Understanding dilution helps ensure that substances are used at proper concentrations, facilitating safe and effective laboratory work and enabling precise titrations of solutions like HCl in the exercise.