0.15g of sodium hydroxide (NaOH) pellets are dissolved in water to make 3.0L of solution. What is the pH of this solution?

M = mols/L

mols = 0.15g/molar mass NaOH.
L = 3.0
pOH = -log(OH^-), then
pH + pOH = pKw = 14
You know pOH and pKw, solve for pH.

To determine the pH of a solution, you need to know the concentration of hydrogen ions (H+) in that solution. To calculate the concentration of H+ ions, you need to know the concentration of hydroxide ions (OH-) since they are related through the equation: Kw = [H+][OH-], where Kw is the ion product constant of water.

Given that the solution contains 0.15g of sodium hydroxide (NaOH) pellets dissolved in 3.0L of water, we need to determine the concentration of OH- ions. We can do this by first calculating the number of moles of NaOH and then using the volume of the solution to find the concentration.

1. Calculate the number of moles of NaOH:
To do this, you need to know the molar mass of NaOH, which can be obtained from the periodic table.
The molar mass of NaOH is:
Na: 22.99 g/mol
O: 16.00 g/mol
H: 1.01 g/mol

NaOH molar mass = 22.99 + 16.00 + 1.01 = 40.00 g/mol

Now, divide the mass of NaOH by its molar mass to find the number of moles:
Number of moles of NaOH = 0.15g / 40.00 g/mol

2. Calculate the concentration of OH- ions:
The concentration of OH- ions is equal to the number of moles of NaOH divided by the volume of the solution.
Concentration of OH- ions = Number of moles of NaOH / Volume of solution

In this case, the volume of the solution is given as 3.0L.

3. Calculate the concentration of H+ ions:
Since the solution is neutral (NaOH is a strong base), the concentration of H+ ions will be equal to the concentration of OH- ions.

4. Calculate the pOH of the solution:
The pOH is calculated using the formula: pOH = -log[OH-]

5. Calculate the pH of the solution:
The pH is calculated using the formula: pH = 14 - pOH

Therefore, to find the pH of the solution, you need to calculate the concentration of OH- ions, then the pOH, and finally the pH using the formulas described above.