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.
Does OOH have a name? It's a new one on me.
Ogre's ocean water apparently
To determine the concentration of sodium, magnesium, and calcium ions in the undiluted OOH (original sample), we need to use the given information about the dilute solutions and titration results.
Let's break down the steps:
Step 1: Calculate the moles of NaOH used in the titration for measuring the total cation concentration.
- The volume of NaOH used is 27.42 mL.
- The concentration of NaOH is 0.1134 M.
- Convert the volume to liters: 27.42 mL ÷ 1000 = 0.02742 L.
- Calculate the moles of NaOH: moles = concentration × volume = 0.1134 M × 0.02742 L = 0.003109 moles of NaOH used.
Step 2: Calculate the moles of EDTA used in the titration for measuring the calcium and magnesium ion concentrations.
- The volume of EDTA used is 34.03 mL.
- The concentration of EDTA is 0.01018 M.
- Convert the volume to liters: 34.03 mL ÷ 1000 = 0.03403 L.
- Calculate the moles of EDTA: moles = concentration × volume = 0.01018 M × 0.03403 L = 0.0003471 moles of EDTA used.
Step 3: Calculate the concentrations of sodium ions, magnesium ions, and calcium ions in the undiluted OOH.
3.1. Sodium ions:
- The moles of NaOH used in the titration (Step 1) represents the moles of sodium ions in the aliquot.
- The volume of the aliquot used is 25.00 mL.
- Convert the volume to liters: 25.00 mL ÷ 1000 = 0.02500 L.
- Calculate the concentration of sodium ions:
concentration (Na+) = moles (NaOH) / volume (aliquot) = 0.003109 moles / 0.02500 L = 0.1244 M.
3.2. Magnesium ions:
- The moles of EDTA used in the titration (Step 2) represents the moles of magnesium ions in the aliquot.
- Since the ratio of magnesium to calcium ions in OOH is 3:2, we can calculate the moles of magnesium ions using this ratio.
- Calculate the moles of magnesium ions:
moles (Mg2+) = (moles (EDTA) × 3) / 2 = (0.0003471 moles × 3) / 2 = 0.00052065 moles of Mg2+ used.
- The volume of the aliquot used is 25.00 mL.
- Convert the volume to liters: 25.00 mL ÷ 1000 = 0.02500 L.
- Calculate the concentration of magnesium ions:
concentration (Mg2+) = moles (Mg2+) / volume (aliquot) = 0.00052065 moles / 0.02500 L = 0.02083 M.
3.3. Calcium ions:
- Since the ratio of magnesium to calcium ions in OOH is 3:2, we can calculate the moles of calcium ions using the moles of magnesium ions calculated in Step 3.2.
- Calculate the moles of calcium ions:
moles (Ca2+) = (moles (Mg2+) × 2) / 3 = (0.00052065 moles × 2) / 3 = 0.0003471 moles of Ca2+ used.
- The volume of the aliquot used is 25.00 mL.
- Convert the volume to liters: 25.00 mL ÷ 1000 = 0.02500 L.
- Calculate the concentration of calcium ions:
concentration (Ca2+) = moles (Ca2+) / volume (aliquot) = 0.0003471 moles / 0.02500 L = 0.01388 M.
So, the concentrations of sodium, magnesium, and calcium ions in the undiluted OOH (original sample) are as follows:
- Sodium ions (Na+): 0.1244 M
- Magnesium ions (Mg2+): 0.02083 M
- Calcium ions (Ca2+): 0.01388 M