A 0.25 g sample of lime (CaO) is dissolved in enough water to make 1487 mL of solution. Calculate the pH of the solution
0.25g/molar mass CaO = moles CaO.
Then M Ca(OH)2 = mols CaO/0l.1487L = ?
Concn OH^- is twice that. I would convert to pOH then to pH.
To calculate the pH of the solution, we need to consider the dissociation of water and the dissociation of calcium oxide (lime) in water.
First, let's find the moles of calcium oxide (lime). We know that the molar mass of CaO is 56.08 g/mol, and we have 0.25 g of CaO. Using this information, we can calculate the number of moles:
moles of CaO = mass / molar mass
moles of CaO = 0.25 g / 56.08 g/mol
moles of CaO = 0.0045 mol
Next, we need to convert the volume of the solution from mL to L. We have 1487 mL of solution:
volume of solution = 1487 mL * (1 L / 1000 mL)
volume of solution = 1.487 L
Now, we can calculate the concentration of calcium oxide (lime) in the solution:
concentration = moles / volume
concentration = 0.0045 mol / 1.487 L
concentration = 0.00303 M
Since calcium oxide (lime) is a strong base, it completely dissociates in water to give calcium ions (Ca2+) and hydroxide ions (OH-). Therefore, the concentration of hydroxide ions is equal to the concentration of calcium oxide.
Now, let's calculate the pOH, which is the negative logarithm of the concentration of hydroxide ions:
pOH = -log[OH-]
pOH = -log(0.00303 M)
pOH ≈ 2.52
Since pH + pOH = 14 at 25°C, we can find the pH of the solution:
pH = 14 - pOH
pH = 14 - 2.52
pH ≈ 11.48
Therefore, the pH of the solution is approximately 11.48.