What mass of HNO3 is needed to prepare 400mL of 0.25N solution?
How many equivalents do you want?
That's N x L = ?
Then equivalents = grams/equivalent weight.
The equivalent weight of HNO3 is the same as the molar mass.
To determine the mass of HNO3 needed to prepare a 0.25N solution, we need to follow a specific calculation process. Here's how you can do it:
Step 1: Convert the given volume from milliliters (mL) to liters (L).
Given volume: 400 mL
Converting to liters: 400 mL ÷ 1000 = 0.4 L
Step 2: Define the concentration of the solution in terms of N (normality).
Given concentration: 0.25N
Step 3: The definition of normality (N) states that 1N is equivalent to 1 gram equivalent weight (GEW) of solute per liter of solution.
So, to determine the mass of HNO3 needed, we need to find the gram equivalent weight of HNO3.
The molecular formula of HNO3 is:
H = 1 atomic mass unit (amu)
N = 14 amu
O = 16 amu
HNO3 has one hydrogen atom, one nitrogen atom, and three oxygen atoms. To calculate the gram molecular weight (GMW) of HNO3, we sum up the atomic masses of the constituent atoms:
GMW = (1 amu × 1) + (14 amu × 1) + (16 amu × 3) = 1 + 14 + 48 = 63 grams
Since HNO3 dissociates into one H+ ion and one NO3- ion, the gram equivalent weight (GEW) is equal to its gram molecular weight (GMW).
Step 4: Use the formula, C₁V₁ = C₂V₂, to calculate the mass of HNO3.
C₁ = Concentration of stock solution (N)
V₁ = Volume of stock solution (L)
C₂ = Concentration of desired solution (N)
V₂ = Volume of desired solution (L)
Given:
C₁ = 0.25N (Concentration of stock solution)
V₁ = ?
C₂ = 0.25N (Concentration of desired solution)
V₂ = 0.4 L (Volume of desired solution)
Rearranging the formula, we get:
V₁ = (C₂ × V₂) ÷ C₁
Substituting the given values:
V₁ = (0.25N × 0.4 L) ÷ 0.25N
V₁ = (0.1 L) ÷ (0.25N/1)
V₁ = 0.1 L ÷ 0.25N
Since 1N = 1 gram equivalent weight (63 grams in our case):
V₁ = 0.1 L ÷ (0.25 × 63 g/1)
V₁ = 0.1 L ÷ 15.75 g
V₁ ≈ 0.006349 g
Therefore, approximately 0.006349 grams of HNO3 is needed to prepare 400 mL of a 0.25N solution.