Question

Calculate the \(\mathrm{pH}\) of a solution made by adding \(2.50 \mathrm{~g}\) of lithium oxide \(\left(\mathrm{Li}_{2} \mathrm{O}\right)\) to enough water to make \(1.500 \mathrm{~L}\) of solution.

Short answer

The pH of the solution made by adding 2.50 g of lithium oxide (Li₂O) to enough water to make 1.500 L of the solution is approximately 13.048.

Step-by-step solution

Calculate the concentration of Li2O in the solution

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 = \( \cfrac{0.08365\thinspace mol}{1.500\thinspace L} = 0.05577\thinspace M \)

Calculate the concentration of hydroxide ions (OH-)

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

Calculate the pH of the solution

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.