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
See your post above.
a.) To prepare a 0.50M solution of HCl from concentrated (12M) reagent, we need to dilute the concentrated solution by adding a specific volume of water.
The formula to calculate the dilution is given by:
M1V1 = M2V2
Where:
M1 = initial concentration of the concentrated solution
V1 = initial volume of the concentrated solution
M2 = final desired concentration
V2 = final desired volume
In this case, M1 = 12M, V1 is unknown, M2 = 0.50M, and V2 = 1.00L (since we want to prepare a 1.00L solution).
Substituting the values into the formula:
12M (V1) = 0.50M (1.00L)
Solving for V1:
V1 = (0.50M x 1.00L) / 12M
V1 = 0.0417 L or 41.7 mL
Therefore, to prepare a 1.00L of 0.50M HCl solution from concentrated (12M) reagent, you would measure 41.7 mL of the concentrated solution and then add water to reach a final volume of 1.00L.
b.) To prepare a 0.50M solution of Sodium Carbonate from the pure solid, we will need to determine the amount of Sodium Carbonate needed and then dissolve it in water to obtain a total volume of 1.00L.
The molar mass of Sodium Carbonate (Na2CO3) is:
Na: 22.99 g/mol x 2 = 45.98 g/mol
C: 12.01 g/mol
O: 16.00 g/mol x 3 = 48.00 g/mol
Total: 45.98 g/mol + 12.01 g/mol + 48.00 g/mol = 105.99 g/mol
To prepare a 0.50M solution, we need to calculate the amount of Sodium Carbonate (in grams) needed:
Mass = molar mass x molarity x volume
Mass = 105.99 g/mol x 0.50 mol/L x 1.00 L
Mass = 52.995 g or approximately 53 g
Therefore, to prepare a 1.00L of 0.50M Sodium Carbonate solution from the pure solid, you would weigh approximately 53 grams of Sodium Carbonate and dissolve it in water to achieve a total volume of 1.00L.