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.
To determine the sodium, magnesium, and calcium ion concentrations in the undiluted OOH solution, we need to use the titration results and apply some calculations.
Let's start by calculating the total cation concentration in the dilute OOH solution:
Total cation concentration (mol/L) = (volume of NaOH used in titration * molarity of NaOH) / volume of sample used in titration
Total cation concentration = (27.42 mL * 0.1134 M) / 0.02500 L
Total cation concentration = 0.12471 M
Since the cation exchange column selectively removes cations other than sodium, we can assume that the concentration of sodium ions in the undiluted OOH solution is the same as in the dilute solution, which is 0.12471 M.
Now, let's determine the concentrations of magnesium and calcium ions in the undiluted OOH solution.
Given that the ratio of magnesium to calcium ions in OOH is about 3:2, we can assume that for every 3 moles of magnesium ions, there are 2 moles of calcium ions.
Let's assume x = moles of magnesium ions in the undiluted OOH solution.
Then, (2/3)x = moles of calcium ions in the undiluted OOH solution.
Now, applying the EDTA titration results:
Moles of EDTA used in titration = (volume of EDTA used in titration * molarity of EDTA)
For magnesium ions:
Moles of EDTA used in titration for magnesium = (34.03 mL * 0.01018 M)
Since 1 mole of EDTA reacts with 1 mole of magnesium ion, we can equate the two moles:
(34.03 mL * 0.01018 M) = x moles of magnesium ions
For the calcium ions:
Moles of EDTA used in titration for calcium = (34.03 mL * 0.01018 M)
Since 1 mole of EDTA reacts with 1 mole of calcium ion, we can equate the two moles:
(34.03 mL * 0.01018 M) = (2/3)x moles of calcium ions
Now, we can solve for x (moles of magnesium ions) and (2/3)x (moles of calcium ions) using the above equations.
We can then convert the moles to concentrations by dividing by the volume of the sample used:
Magnesium ion concentration = (x moles of magnesium ions) / 0.025 L
Calcium ion concentration = ((2/3)x moles of calcium ions) / 0.025 L
Finally, we can substitute the values and calculate the concentrations of magnesium and calcium ions:
Magnesium ion concentration = (34.03 mL * 0.01018 M) / 0.025 L
Calcium ion concentration = (2/3)(34.03 mL * 0.01018 M) / 0.025 L
These calculations will give you the sodium, magnesium, and calcium ion concentrations in the undiluted OOH solution.
To determine the concentrations of sodium (Na+), magnesium (Mg2+), and calcium (Ca2+) ions in the undiluted OOH, you can use the information provided in the question. Let's break down the steps to calculate these concentrations:
1. Calculate the concentration of Na+ ions in the undiluted OOH:
- Volume of NaOH used in titration = 27.42 mL
- Molarity of NaOH = 0.1134 M
- Moles of NaOH = (volume x molarity) = (27.42 mL x 0.1134 M) = 3.1098 mmol
- Since the moles of NaOH equal the moles of Na+ ions, we have 3.1098 mmol of Na+ ions in the 25.00 mL aliquot.
- To find the concentration in the undiluted OOH, we need to consider the dilution factor. The dilution factor is the ratio of the final volume (250.0 mL) to the initial volume (50.00 mL).
- Concentration of Na+ ions = (moles/Vfinal) = (3.1098 mmol/250.0 mL) x 1000 mL/L x (1/5) = 24.88 mmol/L
2. Calculate the concentration of Mg2+ and Ca2+ ions in the undiluted OOH:
- Volume of EDTA used in titration = 34.03 mL
- Molarity of EDTA = 0.01018 M
- Moles of EDTA = (volume x molarity) = (34.03 mL x 0.01018 M) = 0.3466 mmol
- The ratio of Mg2+ to Ca2+ ions in OOH is about 3:2. Let's assume x mmol for Mg2+ and y mmol for Ca2+ in the 25.00 mL aliquot.
- Therefore, we have: x/y = (moles of Mg2+/moles of Ca2+) = (3/2), and (x + y) = 0.3466 mmol.
- Solving these equations, we find that x = 0.20796 mmol and y = 0.13864 mmol.
- To find the concentration in the undiluted OOH, we again need to consider the dilution factor.
- Concentration of Mg2+ ions = (moles/Vfinal) = (0.20796 mmol/250.0 mL) x 1000 mL/L x (1/5) = 1.664 mmol/L
- Concentration of Ca2+ ions = (moles/Vfinal) = (0.13864 mmol/250.0 mL) x 1000 mL/L x (1/5) = 1.109 mmol/L
So, the concentrations of sodium, magnesium, and calcium ions in the undiluted OOH are approximately:
- Sodium (Na+): 24.88 mmol/L
- Magnesium (Mg2+): 1.664 mmol/L
- Calcium (Ca2+): 1.109 mmol/L