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0.65 g of sodium hydroxide ( naoh ) pellets are dissolved in water to make 5.0 l of solution. what is the ph of this solution?

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Final Answer:

The pH of the solution formed by dissolving 0.65 g of sodium hydroxide (NaOH) pellets in 5.0 L of water is approximately 13.0.

Step-by-step explanation:

Sodium hydroxide (NaOH) is a strong base that dissociates completely in water to produce hydroxide ions (OH⁻) and sodium ions (Na⁺). The dissociation reaction of NaOH can be represented as:

NaOH (s) → Na⁺ (aq) + OH⁻ (aq)

Given that 0.65 g of NaOH is dissolved in 5.0 L of water, we first need to calculate the molarity of the NaOH solution:

Molarity (M) = moles of solute / volume of solution in liters

To find the number of moles of NaOH, we use its molar mass:

Molar mass of NaOH = 22.99 g/mol (Na) + 15.999 g/mol (O) + 1.008 g/mol (H) = 39.997 g/mol ≈ 40 g/mol

Number of moles = Mass / Molar mass

Number of moles = 0.65 g / 40 g/mol ≈ 0.01625 mol

Now, plug the values into the formula for molarity:

Molarity = 0.01625 mol / 5.0 L = 0.00325 M

Since NaOH completely dissociates in water, the concentration of hydroxide ions (OH⁻) in the solution is also 0.00325 M. The pH of a solution can be calculated using the equation:

pOH = -log[OH⁻]

pOH = -log(0.00325) ≈ 2.49

pH + pOH = 14 (at 25°C)

pH ≈ 14 - 2.49 ≈ 13.0

Therefore, the pH of the NaOH solution is approximately 13.0, indicating a highly basic solution due to the presence of a significant amount of hydroxide ions.

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