A solution of NaOH(aq) contains 5.4 g of NaOH(s) per 100.0 mL of solution. Calculate the pH and the pOH of the solution at 25 °C.
First, calculate M NaOH.
mols NaOH = grams/molar mass = ?
M = mols/L = mols/0.1 L = ? That gives you the (NaOH) and that is the same as the (OH^-)
Now, pOH = -log(OH)^-
and pH + pOH = pKw = 14. Calculate pH from this.
To calculate the pH and pOH of the solution, we first need to determine the concentration of hydroxide ions (OH-) in the solution. We can then use that concentration to find the pOH, and finally, use the pOH to calculate the pH.
1. Calculate the molarity of the NaOH solution:
Molarity (M) = moles of solute / volume of solution (in liters)
First, convert the mass of NaOH to moles. The molar mass of NaOH is 22.99 g/mol for Na + 16.00 g/mol for O + 1.01 g/mol for H = 39.99 g/mol.
Moles of NaOH = mass of NaOH / molar mass of NaOH
Moles of NaOH = 5.4 g / 39.99 g/mol = 0.135 mol
Next, convert the volume of solution to liters:
Volume of solution = 100.0 mL = 100.0 mL × 1 L / 1000 mL = 0.1 L
Now, calculate the molarity:
Molarity (M) = 0.135 mol / 0.1 L = 1.35 M
2. Calculate the concentration of hydroxide ions (OH-):
Since NaOH is a strong base, it dissociates completely in water to produce one hydroxide ion for each sodium ion.
Therefore, the concentration of OH- is equal to the molarity of NaOH solution:
[OH-] = 1.35 M
3. Calculate the pOH:
pOH = -log10([OH-])
Using [OH-] = 1.35 M:
pOH = -log10(1.35) ≈ 0.87
4. Calculate the pH:
pH + pOH = 14
Therefore, pH = 14 - pOH
pH = 14 - 0.87 ≈ 13.13
Therefore, the pH of the solution is approximately 13.13, and the pOH is approximately 0.87.