The amount of CaCO3 that is equivalent to all polyvalent metal ions in one liter of water.
what is molarity of 100 ppm Mn^2+?
What is the ppm of I^-1 in 0.000100 M of BaI2 ?
Thank you!
To find the amount of CaCO3 equivalent to all polyvalent metal ions in one liter of water, you need to determine the molarity of the metal ions in the water.
To calculate the molarity of a substance, you need to know the molecular weight and the mass or volume of the substance. However, in this case, you have been given the concentration in parts per million (ppm).
The conversion from ppm to molarity depends on the molecular weight of the substance. The steps to convert ppm to molarity are as follows:
Step 1: Determine the molecular weight of the metal ion. The atomic weight of manganese (Mn) is approximately 54.94 g/mol.
Step 2: Calculate the mass of the metal ion in the given concentration. If the concentration is given in ppm, simply multiply the concentration by the total mass of the solution.
For example, if you have 100 ppm Mn2+ in a solution with a volume of 1 liter:
Mass of Mn2+ = (100 ppm) x (1 g/L) = 0.1 g
Step 3: Calculate the moles of the metal ion. Divide the mass of the metal ion by its molecular weight.
Moles of Mn2+ = 0.1 g / 54.94 g/mol ≈ 0.0018 mol
Step 4: Calculate the molarity of the metal ion. Divide the moles of the metal ion by the volume of the solution in liters.
Molarity of Mn2+ = 0.0018 mol / 1 L = 0.0018 M
Therefore, the molarity of 100 ppm Mn2+ is approximately 0.0018 M.
Now, let's move on to the second question:
To determine the ppm of I- in a solution of BaI2, you need to know the molarity of BaI2 and the stoichiometry of the reaction between BaI2 and I-.
The formula for BaI2 shows that it contains two moles of I- ions for every mole of BaI2. Therefore, the number of moles of I- in the solution will be twice the number of moles of BaI2.
To calculate the ppm of I- in the solution:
Step 1: Calculate the moles of BaI2 in the solution using the molarity (M) and volume (V) of the solution. In this case, the concentration is given as 0.000100 M.
Moles of BaI2 = (0.000100 mol/L) x (V L)
Step 2: Calculate the moles of I- in the solution by multiplying the moles of BaI2 by the stoichiometric factor of 2.
Moles of I- = 2 x Moles of BaI2
Step 3: Calculate the mass of I- in the solution using the moles of I- and the molecular weight of I- (126.90 g/mol).
Mass of I- = Moles of I- x Molecular weight of I-
Step 4: Calculate the concentration of I- in parts per million (ppm). This can be done by dividing the mass of I- by the mass of the solution and multiplying by 10^6.
Concentration of I- (ppm) = (Mass of I- / Mass of solution) x 10^6
Please provide the volume of the solution (V L) in order to calculate the ppm of I- in BaI2.