Describe and show your calculations on how you would prepare 1.00L of a 0.50M solution of each of the following
a.) HCl from "concentrated" (12M) reagent.
b.) Sodium Carbonate from the pure solid
a.
c1v1 = c2v2
c = concn
v = volume
Substitute and solve for the one unknown.
b.
You want how many mols? That's mols Na2CO3 = M x L = ?
Then mols = grams/molar mass. You know mols and molar mass; solve for grams Na2CO3.
a.) To prepare 1.00L of a 0.50M solution of HCl from a concentrated (12M) reagent, we need to dilute the concentrated solution with water.
The formula to calculate the volume of concentrated solution required is:
C₁V₁ = C₂V₂
Where:
C₁ = Concentration of the concentrated solution (12M)
V₁ = Volume of the concentrated solution we need to calculate
C₂ = Desired concentration of the diluted solution (0.50M)
V₂ = Desired final volume of the diluted solution (1.00L)
Rearranging the equation, we get:
V₁ = (C₂ * V₂) / C₁
Substituting the values, we have:
V₁ = (0.50M * 1.00L) / 12M
V₁ = 0.0417L
Therefore, to prepare 1.00L of a 0.50M HCl solution, we need to dilute 41.7 mL of the concentrated 12M HCl reagent with water.
b.) To prepare 1.00L of a 0.50M solution of sodium carbonate (Na2CO3) from the pure solid, we need to calculate the amount of the solid required and dissolve it in water.
The formula to calculate the mass of the solid required is:
Mass = (Molarity * Molecular weight * Volume) / 1000
Where:
Molarity = Desired concentration of the solution (0.50M)
Molecular weight = Molecular weight of sodium carbonate (105.99 g/mol)
Volume = Desired final volume of the solution (1.00L)
Substituting the values, we have:
Mass = (0.50M * 105.99 g/mol * 1.00L) / 1000
Mass = 0.052995 g
Therefore, to prepare 1.00L of a 0.50M sodium carbonate solution, we need to dissolve 52.995 mg (or approximately 53 mg) of the pure solid sodium carbonate in water.