What is the pH and pOH of 6.5 g of sodium hydroxide in 650 mL of water?

The question is –

What is the pH and pOH of 6.5 g of sodium hydroxide in 650 mL of water?

Answer:

⇒ The pH is approximately 13.40 and the pOH is approximately 0.60.

Explanation:

First, we’ll calculate the molarity (M) of the sodium hydroxide (NaOH) solution. NaOH is a strong base that completely ionizes in water to produce hydroxide ions (OH-).

Step 1: Convert the mass of NaOH into moles. The molar mass of NaOH is approximately 40 g/mol (23 g/mol for Na, 16 g/mol for O, and 1 g/mol for H).

⇒ 6.5 g NaOH * (1 mol NaOH / 40 g NaOH) = 0.1625 mol NaOH

Step 2: Convert the volume of the solution from mL to L.

⇒ 650 mL * (1 L / 1000 mL) = 0.65 L

Step 3: Calculate the molarity (M) of the solution.

⇒ Molarity = moles of solute/liters of solution

∴ M = 0.1625 mol NaOH / 0.65 L = 0.25 M

Now, let’s find the pOH.

In water, NaOH will dissociate into Na⁺ and OH^– ions. Because it is a strong base, we can assume this dissociation is complete, so the concentration of OH- ions will be equal to the molarity of NaOH, which we found to be 0.25 M.

The pOH is the negative logarithm (base 10) of the concentration of hydroxide ions:

⇒ pOH = -log[OH-]

∴ pOH = -log(0.25) ≈ 0.60

Since pH + pOH = 14 at 25 degrees Celsius, we can find the pH by subtracting our pOH value from 14:

⇒ pH = 14 – pOH

∴ pH = 14 – 0.60 = 13.40

Therefore, for a solution made by dissolving 6.5 g of NaOH in 650 mL of water, the pH is approximately 13.40 and the pOH is approximately 0.60.