How much water must be added to 80 mL of a 2.5N H₂CO₃ solution to achieve a 1.8N solution?

How much water must be added to 80 mL of a 2.5N H₂CO₃ solution to achieve a 1.8N solution?

• A. 10.11 mL
• B. 21.11 mL
• C. 31.11 mL
• D. 41.11 mL

Answer: (C)

To determine how much water must be added to dilute a 2.5N H₂CO₃ solution to a 1.8N solution, you need to use the concept of dilution, where the product of the initial concentration and volume equals the product of the final concentration and volume. First, calculate the number of equivalents of the solute in the initial solution. This is found by multiplying the normality (N) by the volume (V) of the solution: \( \text{Equivalents} = \text{Normality} \times \text{Volume} \) For the original solution: \( \text{Equivalents}_\text{initial} = 2.5 \, \text{N} \times 80 \, \text{mL} = 200 \, \text{equivalents} \) Let’s define the final volume after adding water as \( V_f \). The final concentration (normality) is given as 1.8N. The equation for the final condition is: \( 1.8 \, \text{N} \times V_f = 200 \, \text{equivalents} \) Now, solve for \(

To determine how much water must be added to dilute a 2.5N H₂CO₃ solution to a 1.8N solution, you need to use the concept of dilution, where the product of the initial concentration and volume equals the product of the final concentration and volume.

First, calculate the number of equivalents of the solute in the initial solution. This is found by multiplying the normality (N) by the volume (V) of the solution:

( \text{Equivalents} = \text{Normality} \times \text{Volume} )

For the original solution:

( \text{Equivalents}_\text{initial} = 2.5 , \text{N} \times 80 , \text{mL} = 200 , \text{equivalents} )

Let’s define the final volume after adding water as ( V_f ). The final concentration (normality) is given as 1.8N. The equation for the final condition is:

( 1.8 , \text{N} \times V_f = 200 , \text{equivalents} )

Now, solve for (