A 1N solution of NaOH is diluted 4/20 then rediluted 1:50. What is the final normality? How many grams of NaOH are present in 100ml of the final solution?
I sort of understand the basic concept, but I don't really know how to do normality problems.
This isn't so much a normality problem as it is a dilution (concentration) problem.
If you take 4 ml and dilute to 20 you have diluted it 5 times, right? Then another 50 times in the 1:50. So the total dilution is 5*50 = 250 times. So if you have 1N solution and it's diluted 250 times it is not 1/250 = 0.004N, right? But you can do it with the dilution formula.
c1v1 = c2v2
c = concn
v = volume
1N x 1mL = C2*20 mL
c2 = 1*1/20 = 0.2N after the first dilution. For the second we take 1 and dilute it to 50.
c1v1 = c2v2
0.2N x 1 mL = c2*50 mL
c2 = 0.2*1/50 = 0.004.
So there are 40 g in a liter of 1N solution. There will be 40 x 100/1000 mL = 4g in 100 mL.
To calculate the final normality and the grams of NaOH present in the final solution, let's break down the problem into steps:
Step 1: Understanding Normality
Normality is a measure of the concentration of a solution, specifically the concentration of reactive species in a solution. In the case of NaOH, normality is the concentration of hydroxide ions (OH-).
Step 2: Understanding Dilution
Dilution is a process of decreasing the concentration of a solution by adding more solvent (liquid). This is typically done by adding a known volume of the original solution to a larger volume of solvent.
Step 3: Solving the Problem
a) First, let's calculate the normality after the initial dilution. The original solution is 1N (1 normal). Diluting it by 4/20 means we are taking 4 parts of the original solution and adding 16 parts of solvent. This means we are adding 4/20 (or 1/5) of the original volume, resulting in a 1/5 dilution.
To calculate the normality after the 1/5 dilution, we divide the initial normality by the dilution factor:
Initial normality / Dilution factor = Final normality
Final normality = 1N / 5 = 0.2N
b) Now, we need to calculate the normality after the redilution. The solution is rediluted in a 1:50 ratio, which means 1 part of the 0.2N solution is added to 49 parts of solvent.
Using the same formula as before:
Initial normality / Dilution factor = Final normality
Final normality = 0.2N / 50 = 0.004N
c) To calculate the grams of NaOH in 100ml of the final solution, we need to use the equation:
Grams = Volume x Concentration x Molar Mass
First, we need to find the concentration of NaOH in the final solution. Since the normality is given, we need to convert it to molarity (M), which is the concentration in moles per liter.
To convert normality to molarity for NaOH, we need to multiply the normality (N) by the molar mass of NaOH (40g/mol) because NaOH donates one hydroxide ion (OH-) per mole.
Molarity (M) = Normality (N) × Molar Mass (g/mol)
Molarity = 0.004N × 40g/mol = 0.16M
Now, since we are given that the volume is 100ml (0.1L), we can substitute these values into the equation:
Grams = 0.1L × 0.16M × 40g/mol
Grams = 0.64g
Therefore, there are 0.64 grams of NaOH present in 100ml of the final solution.