50.00 mL of OOH was diluted to 250.0 mL with deionized water. In the determination of the total cation concentration, 25.00 mL aliquots of the dilute OOH were passed through a cation exchange column and the eluent titrated with an average 27.42 mL of 0.1134 M NaOH. In the determination of calcium and magnesium ions, 25.00 mL aliquots of the dilute OOH were titrated with an average 34.03 mL of 0.01018 M EDTA. Determine the sodium, magnesium and calcium ions concentrations in the undilute OOH, noting that the ratio of magnesium to calcium ions in OOH is about 3:2.
Why did the cation go to therapy? Because it had some serious issues with ion-timacy!
Now, let's solve this chemistry puzzle. We have all the information we need to determine the concentrations of the sodium, magnesium, and calcium ions in the undiluted OOH.
First, let's start with the sodium ions. We know that 25.00 mL aliquots of the dilute OOH were titrated with an average of 27.42 mL of 0.1134 M NaOH. This means that 27.42 mL of the NaOH solution is required to neutralize the sodium ions present in each 25.00 mL aliquot.
Using the formula M1V1 = M2V2, where M1 is the concentration of the solution we are trying to find, V1 is the volume of that solution used in the titration (25.00 mL in this case), M2 is the concentration of NaOH, and V2 is the volume of NaOH used (27.42 mL), we can calculate the concentration of sodium ions in the undiluted OOH.
M1 * 25.00 mL = 0.1134 M * 27.42 mL
M1 = (0.1134 M * 27.42 mL) / 25.00 mL
M1 = 0.1239 M
So, the concentration of sodium ions in the undiluted OOH is approximately 0.1239 M.
Now, let's move on to the magnesium and calcium ions. We know that the ratio of magnesium to calcium ions in OOH is about 3:2, and 25.00 mL aliquots of the dilute OOH were titrated with an average of 34.03 mL of 0.01018 M EDTA solution to determine the concentrations of these ions.
Since magnesium and calcium ions have a 3:2 ratio, for every 3 moles of magnesium ions, we have 2 moles of calcium ions. This means that the total number of moles of magnesium and calcium ions in each 25.00 mL aliquot is the same.
Using the formula M1V1 = M2V2, where M1 is the concentration of the solution we are trying to find (the combined magnesium and calcium ions concentration), V1 is the volume of that solution used in the titration (25.00 mL in this case), M2 is the concentration of EDTA, and V2 is the volume of EDTA used (34.03 mL), we can calculate the concentration of the combined magnesium and calcium ions in the undiluted OOH.
M1 * 25.00 mL = 0.01018 M * 34.03 mL
M1 = (0.01018 M * 34.03 mL) / 25.00 mL
M1 = 0.01386 M
Since the ratio of magnesium to calcium ions is 3:2, we can calculate their individual concentrations. Let's assume the concentration of magnesium ions is 3x and the concentration of calcium ions is 2x.
Therefore, we have:
3x + 2x = 0.01386 M
5x = 0.01386 M
x = 0.002772 M
So, the concentration of magnesium ions in the undiluted OOH is approximately 3 * 0.002772 M = 0.008316 M, and the concentration of calcium ions is approximately 2 * 0.002772 M = 0.005544 M.
To summarize:
- Sodium ions concentration in the undiluted OOH: approximately 0.1239 M
- Magnesium ions concentration in the undiluted OOH: approximately 0.008316 M
- Calcium ions concentration in the undiluted OOH: approximately 0.005544 M
To determine the concentrations of sodium, magnesium, and calcium ions in the undiluted OOH, we need to use the information provided about the dilutions and titrations.
Let's break down the steps:
1. Dilution: 50.00 mL of OOH was diluted to 250.0 mL with deionized water. This means that the dilution factor is 250.0 mL / 50.00 mL = 5.
2. Total cation concentration: 25.00 mL aliquots of the diluted OOH were passed through a cation exchange column, and the eluent was titrated with an average of 27.42 mL of 0.1134 M NaOH. Here, NaOH is used to titrate the total cation concentration, so we can assume that the titration reaction is:
OH- + H+ → H2O
The balanced equation tells us that 1 mole of NaOH reacts with 1 mole of H+ ions. From the titration result, we can calculate the concentration of H+ ions in the aliquot.
27.42 mL of 0.1134 M NaOH = (27.42 mL) × (0.1134 mol/L) × (1 mole H+/1 mole NaOH) = 0.3111 mol H+
Since the dilution factor is 5, the concentration of H+ ions in undiluted OOH is:
(0.3111 mol H+)/5 = 0.0622 mol H+/L
3. Calcium and magnesium ions determination: 25.00 mL aliquots of the diluted OOH were titrated with an average of 34.03 mL of 0.01018 M EDTA. The EDTA is used to chelate (bind) with calcium and magnesium ions. The reaction can be written as follows:
Ca2+ + EDTA4- → Ca(EDTA)2-
Mg2+ + EDTA4- → Mg(EDTA)2-
From the stoichiometry of the reaction, we know that 1 mole of calcium or magnesium ion reacts with 1 mole of EDTA. Therefore, the number of moles of calcium and magnesium in the aliquot can be calculated:
34.03 mL of 0.01018 M EDTA = (34.03 mL) × (0.01018 mol/L) = 0.347 mol EDTA
Since we are assuming the ratio of magnesium to calcium ions in OOH is 3:2, we can use this ratio to deduce the concentration of calcium and magnesium ions in the undiluted OOH.
Let's denote the concentration of calcium ions as [Ca2+] and the concentration of magnesium ions as [Mg2+].
From the reaction stoichiometry, we know that 1 mole of EDTA reacts with 1 mole of calcium or magnesium ions. Thus:
0.347 mol EDTA = (0.347 mol Ca2+) + (0.347 mol Mg2+)
Since the ratio of Mg2+ to Ca2+ is 3:2, we can set up the following equation:
(0.347 mol Ca2+) / (0.347 mol Mg2+) = 2/3
Solving for the concentration of calcium ions, we get:
[Ca2+] = (2/3) × [Mg2+]
Finally, using the dilution factor, we can determine the concentrations of sodium, magnesium, and calcium ions in the undiluted OOH:
Sodium (Na+): The concentration of sodium ions remains the same after dilution. Therefore, the concentration of Na+ ions in the undiluted OOH is 0.1134 M.
Magnesium (Mg2+): From the EDTA titration, the concentration of magnesium ions in the aliquot is 0.347 mol / 25.00 mL = 0.01388 mol/mL. However, this is the concentration after dilution. So, the concentration of Mg2+ ions in the undiluted OOH is 0.01388 mol/mL × 5 dilution factor = 0.0694 mol/mL.
Calcium (Ca2+): Using the ratio [Ca2+] = (2/3) × [Mg2+], we can calculate the concentration of calcium ions:
[Ca2+] = (2/3) × 0.0694 mol/mL = 0.04627 mol/mL
Therefore, the concentrations of the sodium, magnesium, and calcium ions in the undiluted OOH are:
Sodium (Na+): 0.1134 M
Magnesium (Mg2+): 0.0694 mol/mL
Calcium (Ca2+): 0.04627 mol/mL
To determine the concentrations of sodium, magnesium, and calcium ions in the undiluted OOH, we need to use the data from the cation exchange column and the titration with EDTA.
First, let's determine the concentration of sodium ions (Na+) in the undiluted OOH.
1. Calculate the moles of NaOH used in the cation exchange column titration:
Moles of NaOH = Volume of NaOH (in L) x Concentration of NaOH (in mol/L)
Moles of NaOH = 27.42 mL x (1 L / 1000 mL) x 0.1134 mol/L
2. Determine the moles of sodium ions in the eluent from the cation exchange column:
Since 25.00 mL of the dilute OOH was passed through the column,
Moles of Na+ = Moles of NaOH / 25.00 mL
3. Calculate the concentration of sodium ions in the undiluted OOH:
Concentration of Na+ = (Moles of Na+ / Volume of OOH used) x Dilution factor
Dilution factor = (Volume of OOH used / Volume of dilute OOH used)
Dilution factor = (250.0 mL / 50.0 mL)
Concentration of Na+ = (Moles of Na+ / 250.0 mL) x (250.0 mL / 50.0 mL)
Next, let's determine the concentrations of magnesium (Mg2+) and calcium (Ca2+) ions in the undiluted OOH using the titration with EDTA.
1. Calculate the moles of EDTA used in the titration:
Moles of EDTA = Volume of EDTA (in L) x Concentration of EDTA (in mol/L)
Moles of EDTA = 34.03 mL x (1 L / 1000 mL) x 0.01018 mol/L
2. Determine the moles of magnesium and calcium ions in the undiluted OOH:
Since 25.00 mL of the dilute OOH was titrated with EDTA,
Moles of Mg2+ + Moles of Ca2+ = Moles of EDTA / 25.00 mL
3. Use the given 3:2 ratio of Mg2+ to Ca2+ to calculate their individual moles:
Let Moles of Mg2+ be 3x and Moles of Ca2+ be 2x,
3x + 2x = Moles of EDTA / 25.00 mL
Solve for x, which will give us the moles of calcium ions.
4. Calculate the concentrations of magnesium and calcium ions in the undiluted OOH:
Concentration of Mg2+ = (3x / 250.0 mL) x (250.0 mL / 50.0 mL)
Concentration of Ca2+ = (2x / 250.0 mL) x (250.0 mL / 50.0 mL)
Remember to convert the volumes from mL to L and perform the necessary unit conversions to ensure consistent units throughout the calculations.
Note: The above calculations assume that all other cations present in the OOH do not interfere with the cation exchange column or the EDTA titration.