Calculate the pH of a solution made by adding 2.40 g of lithium oxide to enough water to make 1.300 L of solution.
Convert 2.40 g Li2O to moles. moles = grams/molar mass
Then M = mols/L soln.
To calculate the pH of a solution, we need to determine the concentration of hydrogen ions (H+) in the solution. In this case, we can use the dissociation reaction of lithium oxide (Li2O) in water to find the concentration of hydroxide ions (OH-) since water is amphoteric.
The dissociation reaction of lithium oxide in water is:
Li2O + H2O → 2Li+ + 2OH-
First, we need to determine the number of moles of lithium oxide:
Molar mass of Li2O = molar mass of Li + molar mass of O = 2(6.94 g/mol) + 16.00 g/mol = 29.88 g/mol
Number of moles of Li2O = mass of Li2O / molar mass of Li2O = 2.40 g / 29.88 g/mol ≈ 0.0802 mol
The number of moles of OH- ions produced will be twice the number of moles of Li2O because of the balanced equation:
Number of moles of OH- ions = 2 × number of moles of Li2O = 2 × 0.0802 mol = 0.1604 mol
Next, we need to find the concentration of hydroxide ions in the solution:
Concentration of OH- ions = number of moles of OH- ions / volume of solution
Volume of solution = 1.300 L
Concentration of OH- ions = 0.1604 mol / 1.300 L ≈ 0.1237 M
Since water is amphoteric, the concentration of hydroxide ions (OH-) is equal to the concentration of hydrogen ions (H+). Therefore, the concentration of hydrogen ions is also 0.1237 M.
Finally, we can calculate the pH of the solution using the formula:
pH = -log[H+]
pH = -log(0.1237) ≈ 0.91
Therefore, the pH of the solution made by adding 2.40 g of lithium oxide to enough water to make 1.300 L is approximately 0.91.