A 0.56-g sample of barium hydroxide, Ba(OH)2 (molar mass 171.35 g/mol), is dissolved in enough water to make 250.0 mL of solution. What is the pH of the resulting solution?
(A) pH = 1.58 (B) pH- 1.88 (C) pH 12.12 (D) pH 12.42
The correct answer and explanation is :
To calculate the pH of the resulting solution, we need to consider the dissociation of barium hydroxide (Ba(OH)₂) in water. Ba(OH)₂ is a strong base, so it dissociates completely in water into barium ions (Ba²⁺) and hydroxide ions (OH⁻).
The dissociation reaction is:
[
\text{Ba(OH)}_2 (s) \rightarrow \text{Ba}^{2+} (aq) + 2 \text{OH}^- (aq)
]
Step 1: Calculate the moles of Ba(OH)₂
We are given that the sample of Ba(OH)₂ weighs 0.56 g, and the molar mass of Ba(OH)₂ is 171.35 g/mol. To find the moles of Ba(OH)₂:
[
\text{moles of Ba(OH)}_2 = \frac{\text{mass of Ba(OH)}_2}{\text{molar mass of Ba(OH)}_2}
]
[
\text{moles of Ba(OH)}_2 = \frac{0.56 \, \text{g}}{171.35 \, \text{g/mol}} = 0.00327 \, \text{mol}
]
Step 2: Determine the concentration of OH⁻
Since Ba(OH)₂ dissociates completely, for every mole of Ba(OH)₂, we get 2 moles of OH⁻. Thus, the number of moles of OH⁻ is:
[
\text{moles of OH}^- = 2 \times \text{moles of Ba(OH)}_2 = 2 \times 0.00327 \, \text{mol} = 0.00654 \, \text{mol}
]
Now, we calculate the concentration of OH⁻ in the solution. The volume of the solution is 250.0 mL (0.2500 L):
[
\text{[OH}^-] = \frac{\text{moles of OH}^-}{\text{volume of solution in L}} = \frac{0.00654 \, \text{mol}}{0.2500 \, \text{L}} = 0.02616 \, \text{M}
]
Step 3: Calculate the pOH
The pOH is calculated using the formula:
[
\text{pOH} = -\log[\text{OH}^-]
]
[
\text{pOH} = -\log(0.02616) = 1.58
]
Step 4: Calculate the pH
Finally, the pH can be found using the relationship between pH and pOH:
[
\text{pH} + \text{pOH} = 14
]
[
\text{pH} = 14 – \text{pOH} = 14 – 1.58 = 12.42
]
Conclusion:
The pH of the solution is 12.42, so the correct answer is (D) pH = 12.42.