A chemist must prepare 650.0mL of sodium hydroxide solution with a pH of 13.90 at 25°C. He will do this in three steps:

1. Weigh 3.27 ×10⁻⁴ ng of NaOH  in a special balance for weighing in nanograms

2. Measure a volume of 650 mL in a test tube

3. Make the solution by mixture

Explanation:

NaOH is a strong base. As every strong base, can be completely dissociated:

NaOH →  Na⁺  +  OH⁻

If pH is 13.90, we can calculate the [OH⁻], so we would know the molarity of NaOH.

1 mol of NaOH can generate 1 mol of OH⁻

[OH⁻] = 10⁻¹³'⁹ = 1.26×10⁻¹⁴ M

In conclussion, molarity of NaOH is 1.26×10⁻¹⁴ M

M = moles of NaOH / volume of solution

We convert the volume from mL to L

650 mL . 1 L /1000mL = 0.650 L

Moles of NaOH = 0.650 L . 1.26×10⁻¹⁴ M = 8.18×10⁻¹⁵ moles

We convert the moles to mass, to determine the grams:

8.18×10⁻¹⁵ moles .  40 g/mol = 3.27 ×10⁻¹³ g

We can convert the grams to ng = 3.27 ×10⁻⁴ ng

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