91Ó°ÊÓ

To find the concentration of Li2O in the solution, we need to know the moles of the compound present in 1.500 L of water. To calculate the moles of Li2O, we'll use the mass (2.50 g) and molar mass of the compound. The molar mass of Li2O is approximately: \(Li_2O = 2 \times 6.939 \thinspace g/mol (mass \thinspace of \thinspace Li)\ +\ 15.999 \thinspace g/mol (mass \thinspace of \thinspace O) = 29.877 \thinspace g/mol\) Now, we can find the moles of Li2O using the given mass: moles of Li2O = \( \cfrac{mass}{molar\thinspace mass} = \cfrac{2.50\thinspace g}{29.877\thinspace g/mol} = 0.08365\thinspace mol \) Now that we have the moles, we can calculate the concentration (in mol/L) by dividing the moles by the volume of the solution: Concentration of Li2O = \( \cfrac{0.08365\thinspace mol}{1.500\thinspace L} = 0.05577\thinspace M \)

When Li2O dissolves in water, it reacts with water to form lithium hydroxide (LiOH), a strong base: \(Li_2O + H_2O \longrightarrow 2LiOH\) Notice that the reaction has a 1:2 molar ratio between Li2O and LiOH. Therefore, we need to find the concentration of hydroxide ions (OH-) by multiplying the concentration of Li2O by 2. Concentration of OH- = 2 × (concentration of Li2O) = 2 × 0.05577 M = 0.1115 M

Since LiOH is a strong base, we can calculate the pOH first, and then find the pH. To calculate the pOH, we use the formula: pOH = -log10(concentration of OH-) pOH = -log10(0.1115 M) = 0.9521 Now, we use the relationship between pH and pOH: pH + pOH = 14 pH = 14 - pOH = 14 - 0.9521 = 13.048 Thus, the pH of the solution is approximately 13.048.