Which formula should be used to prepare a 0.3 N NaOH solution from a 5 N NaOH solution?

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Multiple Choice

Which formula should be used to prepare a 0.3 N NaOH solution from a 5 N NaOH solution?

Explanation:
In dilution, the amount of reactive units (equivalents) stays the same. For NaOH, a strong monoprotic base, normality equals molarity, so normality is a convenient way to track how much reactive base you have. Thus the dilution relationship N1V1 = N2V2 applies: the product of normality and volume is constant before and after dilution. To use it, rearrange to find the volume of stock solution needed: V1 = (N2/N1) × V2. With a 5 N stock and aiming for 0.3 N in a final volume V2, V1 = (0.3/5) × V2 = 0.06 × V2. For example, to prepare 1000 mL of 0.3 N, take 60 mL of the 5 N stock and dilute to 1000 mL with water. Other formulas aren’t suited here: PV = nRT is for gases, and P1V1 = P2V2 likewise applies to gas scenarios. The simple definition M = moles/L describes concentration but doesn’t connect how much stock to take to reach a target concentration and volume without the dilution relationship.

In dilution, the amount of reactive units (equivalents) stays the same. For NaOH, a strong monoprotic base, normality equals molarity, so normality is a convenient way to track how much reactive base you have. Thus the dilution relationship N1V1 = N2V2 applies: the product of normality and volume is constant before and after dilution.

To use it, rearrange to find the volume of stock solution needed: V1 = (N2/N1) × V2. With a 5 N stock and aiming for 0.3 N in a final volume V2, V1 = (0.3/5) × V2 = 0.06 × V2. For example, to prepare 1000 mL of 0.3 N, take 60 mL of the 5 N stock and dilute to 1000 mL with water.

Other formulas aren’t suited here: PV = nRT is for gases, and P1V1 = P2V2 likewise applies to gas scenarios. The simple definition M = moles/L describes concentration but doesn’t connect how much stock to take to reach a target concentration and volume without the dilution relationship.

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