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 to enough water to make 1.500 L of solution is approximately \(13.05\).

Step-by-step solution

Write the chemical equation for the reaction of Li2O with water

The chemical equation for the reaction of lithium oxide with water is given below: \[ \mathrm{Li}_{2}\mathrm{O(s)} + \mathrm{H}_{2}\mathrm{O(l)} \longrightarrow 2\mathrm{LiOH(aq)} \] In this reaction, each mole of Li2O reacts with water to produce 2 moles of lithium hydroxide (LiOH), which is a strong base and dissociates completely in water.

Calculate the number of moles of Li2O added to the water

First, let's find the molar mass of Li2O: Molar mass of Li2O = 2×(Molar mass of Li) + Molar mass of O = 2×(6.94 g/mol) + 16.00 g/mol = 29.88 g/mol Now, we can calculate the number of moles of Li2O: Number of moles (n) = mass / molar mass n(Li2O) = 2.50 g / 29.88 g/mol = 0.0837 mol

Determine the moles of OH- ions that will form from the moles of Li2O in the solution

Since one mole of Li2O produces two moles of LiOH, and each mole of LiOH produces one mole of OH- ions, the number of moles of OH- ions in the solution would be: n(OH-) = 2 × n(Li2O) = 2 × 0.0837 mol = 0.1674 mol

Calculate the concentration of OH- ions

Now, we can calculate the concentration of OH- ions in the 1.500 L solution: Concentration (C) = moles/volume C(OH-) = 0.1674 mol / 1.500 L = 0.1116 M

Calculate the pOH of the solution

Now, we will calculate the pOH by taking the negative base-10 logarithm of the OH- concentration: pOH = -log10(C(OH-)) pOH = -log10(0.1116) = 0.951

Determine the pH of the solution by using the relation between pH and pOH

To calculate the pH of the solution, we need to use the relationship between pH and pOH: pH + pOH = 14 Now, we can find the pH: pH = 14 - pOH = 14 - 0.951 = 13.049 The pH of the solution is approximately 13.05.