I need help with this problem:
if you want to reduce the strength of 100 mL of 10 N NaOH to 3N, how much water would you need to add?
Thanks in advance.
mL1 x N1 = mL2 x N2
Oh yeah! Thanks, so you would substitute:
(10)(100) = (3)(x) to get:
x = 333.33 mL, right?
To solve this problem, you will need to understand the concept of normality (N) and how to calculate dilutions.
Normality (N) is a measure of the concentration of a solution, specifically the number of equivalents of a solute per liter of solution. In this case, NaOH is the solute.
To calculate the amount of water needed to dilute a solution, you can use the formula:
(C1)(V1) = (C2)(V2)
Where:
C1 = initial concentration of the solution in N (10 N in this case)
V1 = initial volume of the solution (100 mL in this case)
C2 = final concentration of the solution in N (3 N in this case)
V2 = final volume of the solution (100 mL + volume of water added)
In this problem, you are trying to find the volume of water that needs to be added (V2). Rearranging the formula, we have:
V2 = [(C1)(V1)] / C2
Substituting the values given in the problem:
V2 = [(10 N)(100 mL)] / 3 N
Simplifying the equation:
V2 = 333.33 mL
Therefore, you would need to add approximately 333.33 mL of water to reduce the strength of 100 mL of 10 N NaOH to 3 N.