describe the preparation of 12 mL of a 0.050 M solution of each of the halogens, using the reactions in the preceding question. Available to you are solid Ca(OCl)2; 0.20 M solutions of NaCl, NaBr, NaBrO3, and CuSO4; a 0.40 M solution of NaI; a 0.50 M solution of H2SO4; distilled water; and a 10-mL graduated cylinder (where volumes can be read to the nearest 0.1 mL). Calculate the required quantity of each reagent and the volume of water that must be added for dilution.
a. Chlorine water
b. Bromine water.
c. Iodine water
What I need help with is how do I solve for this question? Do I use dilution calculations to solve for a,b,c?
What's the preceding question?
It's the reactions:
6H++ BrO3-+ 5Br- --> 3H2O+ 3Br2
2 Cu2+ 4I--> 2CuI + I2
2H++OCl- +Cl- --> Cl2+H2O
Yes, you will use dilution calculations to solve for parts a, b, and c. The overall strategy is to calculate the amount of solute needed to prepare the desired solution concentration, and then find the volume of water required for dilution.
Let's break down each part of the question:
a. To prepare 12 mL of a 0.050 M chlorine water solution, you need to calculate the amount of solid Ca(OCl)2 needed and the volume of water required for dilution.
1. Calculate the moles of Ca(OCl)2 needed:
Since the molar ratio between Ca(OCl)2 and Cl2 is 1:2, you need half the moles of Cl2. So:
Moles of Cl2 = Moles of Ca(OCl)2 / 2
Moles of Cl2 = (0.050 mol/L) * (12 mL / 1000 mL) = 0.0006 mol
2. Calculate the mass of Ca(OCl)2 needed:
Use the molar mass of Ca(OCl)2 to find the mass:
Mass of Ca(OCl)2 = Moles of Ca(OCl)2 * Molar mass of Ca(OCl)2
3. Calculate the volume of water for dilution:
Volume of Water = Total Volume - Volume of Solute
Volume of Water = 12 mL - Volume of Ca(OCl)2
b. To prepare 12 mL of a 0.050 M bromine water solution, you will use a similar procedure as in part a, but with a different starting material.
1. Calculate the moles of Br2 needed:
The molar ratio between NaBr and Br2 is 1:1, so the moles of Br2 needed are equal to the desired concentration.
Moles of Br2 = 0.050 mol/L * (12 mL / 1000 mL) = 0.0006 mol
2. Calculate the mass of NaBr needed:
Use the molar mass of NaBr to find the mass:
Mass of NaBr = Moles of NaBr * Molar mass of NaBr
3. Calculate the volume of water for dilution:
Volume of Water = Total Volume - Volume of Solute
Volume of Water = 12 mL - Volume of NaBr
c. For iodine water, you will use a different approach since you have a stock solution of NaI available.
1. Calculate the moles of NaI needed:
The molar ratio between NaI and I2 is 1:1, so the moles of I2 needed are equal to the desired concentration.
Moles of I2 = 0.050 mol/L * (12 mL / 1000 mL) = 0.0006 mol
2. Calculate the volume of NaI solution needed:
Volume of NaI solution = Moles of NaI / Molarity of NaI solution
3. Calculate the volume of water for dilution:
Volume of Water = Total Volume - Volume of NaI solution
Remember to use the appropriate molar masses and molarities to perform the calculations accurately.
Yes, you can use dilution calculations to solve this question. To prepare each of the halogen solutions, you will need to calculate the required quantities of the reagents and determine the volume of water that must be added for dilution.
Let's go step by step for each halogen solution:
a. Chlorine water:
Chlorine water can be prepared by reacting solid Ca(OCl)2 with H2SO4. From the information given, we have a 0.50 M solution of H2SO4. To calculate the required quantity of Ca(OCl)2, we can use the stoichiometry of the reaction and the desired molarity.
The balanced equation for the reaction between Ca(OCl)2 and H2SO4 is:
Ca(OCl)2 + H2SO4 -> CaSO4 + HClO
From the equation, we can see that the ratio of Ca(OCl)2 to HClO is 1:1.
Given that you need a 0.050 M HClO solution, you will also need 0.050 mol/L of Ca(OCl)2 to achieve the desired concentration.
To calculate the required quantity of Ca(OCl)2, use the following formula:
Volume (in L) of Ca(OCl)2 = (moles of Ca(OCl)2) / (molarity of Ca(OCl)2)
Volume = 0.050 mol / 0.050 mol/L = 1 L
Since the question asks for 12 mL of the solution, you will need to measure 12 mL of the 1 L solution and then dilute it with distilled water to a total volume of 12 mL.
b. Bromine water:
Bromine water can be prepared by reacting NaBrO3 with H2SO4. From the information given, we have a 0.50 M solution of H2SO4 and a 0.40 M solution of NaI. To calculate the required quantities of NaBrO3 and NaI, use stoichiometry and the desired molarity.
The balanced equation for the reaction between NaBrO3 and H2SO4 is:
NaBrO3 + 5H2SO4 -> HBrO + NaHSO4 + 2H2O + 3SO2
From the equation, we can see that the ratio of NaBrO3 to HBrO is 1:1.
Given that you need a 0.050 M HBrO solution, you will also need 0.050 mol/L of NaBrO3 to achieve the desired concentration.
To calculate the required quantity of NaBrO3, use the formula mentioned earlier:
Volume (in L) of NaBrO3 = (moles of NaBrO3) / (molarity of NaBrO3)
Volume = 0.050 mol / 0.050 mol/L = 1 L
Since the question asks for 12 mL of the solution, measure 12 mL of the 1 L solution and then dilute it with distilled water to a total volume of 12 mL.
c. Iodine water:
Iodine water can be prepared by reacting NaI with CuSO4. From the information given, we have a 0.20 M solution of NaBr and a 0.40 M solution of NaI. To calculate the required quantities of NaI, use stoichiometry and the desired molarity.
The balanced equation for the reaction between NaI and CuSO4 is:
2NaI + CuSO4 -> CuI2 + Na2SO4
From the equation, we can see that the ratio of NaI to CuI2 is 2:1.
Given that you need a 0.050 M CuI2 solution, you will also need 0.025 mol/L of NaI to achieve the desired concentration.
To calculate the required quantity of NaI, use the formula mentioned earlier:
Volume (in L) of NaI = (moles of NaI) / (molarity of NaI)
Volume = 0.025 mol / 0.40 mol/L = 0.0625 L
Since the question asks for 12 mL of the solution, measure 12 mL of the 0.0625 L solution and then dilute it with distilled water to a total volume of 12 mL.
Remember to always double-check your calculations and units to ensure accurate results.