Calculate the pH of a solution made by adding 3.50 g of lithium oxide (Li2O) to enough water to make 1.290 L of solution. (Kw = 1.0 10-14)

Why did the scarecrow become a chemist? Because he wanted to be outstanding in his field! But let's get back to your question.

To calculate the pH of the solution, we first need to determine the concentration of hydroxide ions (OH-) in the solution.

Given that 3.50 g of Li2O has been dissolved, we can calculate the moles of Li2O using its molar mass. The molar mass of Li2O is approximately 29.88 g/mol.

Number of moles of Li2O = (3.50 g) / (29.88 g/mol) = 0.117 mol

Since Li2O completely dissociates in water to form two moles of Li+ ions and one mole of O2- ions, the concentration of hydroxide ions can be determined as follows:

Concentration of OH- ions = 2 * (0.117 mol) / (1.290 L) = 0.181 M

Now, using the autoionization constant of water (Kw = 1.0 x 10^-14), we can determine the concentration of hydrogen ions (H+) in the solution.

Kw = [H+][OH-]

[H+] = (Kw) / [OH-] = (1.0 x 10^-14) / (0.181 M) ≈ 5.52 x 10^-14 M

Finally, to calculate the pH, we can use the formula:

pH = -log[H+]

pH = -log(5.52 x 10^-14) ≈ 13.258

So, the pH of the solution is approximately 13.258.

To calculate the pH of a solution, we need to determine the concentration of hydroxide ions (OH-) and use it to calculate the pOH. Then, we can use the equation pH + pOH = 14 to find the pH.

Step 1: Determine the amount of hydroxide ions (OH-) produced by the dissociation of lithium oxide.

The molecular weight of Li2O is:
Li: 6.94 g/mol
O: 16.00 g/mol
So, the molar mass of Li2O is 6.94 g/mol + 16.00 g/mol = 22.94 g/mol.

We have 3.50 g of Li2O, so we can calculate the number of moles of Li2O:
Moles of Li2O = Mass / Molar mass = 3.50 g / 22.94 g/mol = 0.152 mol.

Lithium oxide dissociates in water to produce two lithium ions (Li+) and one hydroxide ion (OH-). Therefore, the number of moles of hydroxide ions produced is also 0.152 mol.

Step 2: Determine the concentration of hydroxide ions (OH-) in the solution.

We have 1.290 L of solution. To calculate the concentration, we divide the number of moles of hydroxide ions by the volume of the solution:
Concentration of OH- = Moles / Volume of solution = 0.152 mol / 1.290 L = 0.118 M.

Step 3: Calculate the pOH.

The pOH is calculated using the formula:
pOH = -log10(OH- concentration).

Plugging in the values, we get:
pOH = -log10(0.118) = 0.927.

Step 4: Calculate the pH.

We can use the equation pH + pOH = 14 to find the pH:
pH = 14 - pOH = 14 - 0.927 = 13.073.

Therefore, the pH of the solution made by adding 3.50 g of lithium oxide (Li2O) is 13.073.

To calculate the pH of the solution, we will need to determine the concentration of hydroxide ions (OH-) in the solution, as pH is a measure of the concentration of hydrogen ions (H+).

1. First, we need to calculate the number of moles of lithium oxide (Li2O) added to the solution. We can do this using the molar mass of Li2O. The molar mass of Li2O is 29.88 g/mol.

Number of moles = Mass / Molar mass
= 3.50 g / 29.88 g/mol
= 0.117 mol

2. Next, we need to determine the molarity (concentration) of hydroxide ions (OH-) in the solution. Since lithium oxide is an ionic compound, it will dissociate completely in water. Each mole of Li2O will produce 2 moles of hydroxide ions (OH-).

Molarity of OH- ions = Number of moles / Volume of solution
= 0.117 mol / 1.290 L
= 0.091 mol/L

3. Next, we need to calculate the concentration of hydrogen ions (H+) in the solution. Since we know that water dissociates into equal concentrations of H+ and OH-, we can use the value of Kw (1.0 x 10^-14) to determine the concentration of H+.

Kw = [H+][OH-]
[H+] = Kw / [OH-]
= (1.0 x 10^-14) / (0.091 mol/L)
= 1.1 x 10^-13 mol/L

4. Finally, we can calculate the pH of the solution using the equation:

pH = -log[H+]
= -log(1.1 x 10^-13)
≈ 12.96

Therefore, the pH of the solution made by adding 3.50 g of lithium oxide to enough water to make 1.290 L of solution is approximately 12.96.

..........Li2O + HOH ==> 2LiOH

1 mol Li2O produces 2 mols LiOH. How many moles Li2O do you have? I see mol = grams/molar mass
mol Li2O = 3.50 g Li2O/molar mass Li2O = ?
Then twice that gives moles LiOH produced.
M = mols LiOH/L LiOH.
M LiOH = moles OH^-; convert to H^+ by
H^+)(OH^-) = Kw ; then convert to pH.
pH = -log(H^+)