Problem 14 - Chemistry and Chemical Reactivity

Question

The \(\mathrm{pH}\) of a solution of \(\mathrm{Ba}(\mathrm{OH})_{2}\) is 10.66 at \(25^{\circ} \mathrm{C} .\) What is the hydroxide ion concentration in the solution? If the solution volume is \(125 \mathrm{mL}\) what mass of \(\mathrm{Ba}(\mathrm{OH})_{2}\) must have been dissolved?

Step-by-Step Solution

Answer

Hydroxide ion concentration is \(4.57 \times 10^{-4} \text{ mol/L}\) and mass of \(\text{Ba(OH)}_2\) is approximately 0.00490 g.

Step 1: Find pOH from pH

We know that the sum of the pH and pOH of a solution at 25°C is 14. Therefore, given that \(\text{pH} = 10.66\), we find the pOH as follows:\[ \text{pOH} = 14 - \text{pH} = 14 - 10.66 = 3.34 \]

Step 2: Calculate the Hydroxide Ion Concentration

We can find the hydroxide ion concentration \([\text{OH}^-]\) using the pOH value. The formula is:\[ [\text{OH}^-] = 10^{-\text{pOH}} \]Substitute the pOH value:\[ [\text{OH}^-] = 10^{-3.34} \approx 4.57 \times 10^{-4} \, \text{mol/L} \]

Step 3: Calculate Moles of OH⁻

Given the volume of the solution is 125 mL, we first convert this volume into liters (since concentration is in mol/L):\[ 125 \, \text{mL} = 0.125 \, \text{L} \]Now calculate the moles of \(\text{OH}^-\) by using the concentration and the volume of the solution:\[ \text{Moles of } \text{OH}^- = \text{Concentration} \times \text{Volume} = 4.57 \times 10^{-4} \, \text{mol/L} \times 0.125 \, \text{L} \approx 5.71 \times 10^{-5} \, \text{mol} \]

Step 4: Determine Moles of Ba(OH)₂ Dissolved

Since \(\text{Ba(OH)}_2\) dissociates into one \(\text{Ba}^{2+}\) ion and two \(\text{OH}^-\) ions per formula unit, the number of moles of \(\text{OH}^-\) ions is twice that of \(\text{Ba(OH)}_2\). Therefore, to find the moles of \(\text{Ba(OH)}_2\), divide the moles of \(\text{OH}^-\) by 2:\[ \text{Moles of Ba(OH)}_2 = \frac{5.71 \times 10^{-5} \, \text{mol}}{2} \approx 2.86 \times 10^{-5} \, \text{mol} \]

Step 5: Calculate Mass of Ba(OH)₂ Dissolved

To find the mass, use the molar mass of \(\text{Ba(OH)}_2\), which is approximately 171.34 g/mol. Use the formula:\[ \text{Mass} = \text{Moles} \times \text{Molar Mass} \]\[ \text{Mass of Ba(OH)}_2 = 2.86 \times 10^{-5} \, \text{mol} \times 171.34 \, \text{g/mol} \approx 0.00490 \, \text{g} \]

Key Concepts

Hydroxide Ion Concentration

To find the hydroxide ion concentration, we need to first understand its relationship with pH and pOH. At 25°C, the sum of pH and pOH is always 14. By knowing the pH, you can easily determine the pOH using the formula:

• \[ \text{pOH} = 14 - \text{pH} \]

Once you have the pOH, the hydroxide ion concentration \( [\text{OH}^-] \) can be calculated with this equation:

• \[ [\text{OH}^-] = 10^{-\text{pOH}} \]

For a pH of 10.66, the pOH is 3.34. By substituting this back into the equation, the hydroxide ion concentration becomes approximately \( 4.57 \times 10^{-4} \) \( \text{mol/L} \). This tells us how many hydroxide ions are present per liter of solution.Understanding these relationships helps in solving various pH and pOH calculations efficiently.

Dissociation of Ba(OH)₂

Barium hydroxide, \( \text{Ba(OH)}_2 \), is a strong base and fully dissociates in water. This means that each unit of \( \text{Ba(OH)}_2 \) dissolves to form one \( \text{Ba}^{2+} \) ion and two \( \text{OH}^- \) ions. This 1:2 ratio is crucial in concentration calculations. When you know the concentration of \( \text{OH}^- \) ions, it impacts how you determine the amount of \( \text{Ba(OH)}_2 \) that was initially dissolved.

• The moles of \( \text{OH}^- \) ions are twice the moles of \( \text{Ba(OH)}_2 \)
• Therefore, to find the moles of \( \text{Ba(OH)}_2 \), we need to divide the moles of \( \text{OH}^- \) by 2

For example, with a calculated 5.71 \( \times 10^{-5} \) moles of \( \text{OH}^- \), there are 2.86 \( \times 10^{-5} \) moles of \( \text{Ba(OH)}_2 \).This clear understanding of dissociation allows for precise calculations of reactants.

Molar Mass

The molar mass of a compound, like \( \text{Ba(OH)}_2 \), is crucial for converting moles into mass. Molar mass is simply the sum of the atomic masses of its constituent elements. For \( \text{Ba(OH)}_2 \), this includes:

• Barium (Ba): approximately 137.33 g/mol
• Oxygen (O): approximately 16.00 g/mol
• Hydrogen (H): approximately 1.01 g/mol

Combined, \( \text{Ba(OH)}_2 \) has a molar mass of about 171.34 g/mol. This value is used to calculate the actual mass of \( \text{Ba(OH)}_2 \) dissolved, by multiplying it by the moles present:

• \[ \text{Mass} = \text{Moles} \times \text{Molar Mass} \]

For the example in our problem, 2.86 \( \times 10^{-5} \) moles translates to approximately 0.00490 g of \( \text{Ba(OH)}_2 \).This step is essential to relate chemical calculations to practical laboratory measurements.