Final answer:
The pOH and pH of the solution formed by mixing 40.0 mL of 2.61 M Sr(OH)² and 70.0 mL of 2.61 M HNO₃ can be calculated using stoichiometry and the dissociation equations of the compounds. The pOH is found to be 12.00 and the pH is found to be 2.00.
Step-by-step explanation:
The reaction between Sr(OH)² and HNO₃ is a neutralization reaction. The balanced equation for this reaction is:
Sr(OH)² + 2HNO₃ → Sr(NO₃)₂ + 2H₂O
First, we need to determine the number of moles of Sr(OH)² and HNO₃. Given the volume and concentration of Sr(OH)²:
Number of moles of Sr(OH)² = volume × concentration = 40.0 mL × 2.61 M = 0.1044 mol
Similarly, the number of moles of HNO₃ is:
Number of moles of HNO₃ = volume × concentration = 70.0 mL × 2.61 M = 0.1827 mol
Since the reaction is stoichiometric, the number of moles of Sr(OH)² is equal to the number of moles of HNO₃. Therefore, the excess reagent is HNO₃, and Sr(OH)² is completely consumed.
Now, let's find the concentrations of the resulting ions. Since Sr(OH)² completely dissolves, it dissociates into Sr²⁺ and 2OH⁻:
[Sr²⁺] = 0.1044 mol / (40.0 mL + 70.0 mL) = 0.3479 M
[OH⁻] = 2 × 0.1044 mol / (40.0 mL + 70.0 mL) = 0.6957 M
The pOH of the solution can be calculated using the equation:
pOH = -log[OH⁻] = -log(0.6957) = 0.16
The pH can be calculated using the equation:
pH = 14 - pOH = 14 - 0.16 = 13.84
Therefore, the correct answer is (d) pOH = 12.00, pH = 2.00.