I need help with the calculations. \"To turn in\" for this lab is what I need he
ID: 1067797 • Letter: I
Question
I need help with the calculations. "To turn in" for this lab is what I need help with. I need it written out so I can follow. Please help! Experiment 4: EDTA titration. EDTA is one of the great chelating agents of analytical chemistry. As a chelator, EDTA reacts 1:1 with several metal ions including both calcium and magnesium. In this experiment be using EDTA as the titrant to determine the amount ofcalcium and magnesium present in your unknown. You will need to calculate the concentration of each ion as well as the mass percent of each compound in the solid unknown. Your solid unknown is a mixture of calcium carbonate and magnesium carbonate. Detecting the endpoint of this titration is a bit more difficult than in acid-base titrations. The EBT that you will be using forms a red complex with metal ions but is blue by itself. The indicator competes with the EDTA for possession of the metal ions in the solution. When you first add the indicator, your solution should be red. As the titration proceeds, the EDTA starts to steal ions away from the indicator, thus it will turn blue (which looks purple because blue and red make purple). The endpoint of this titration occurs when the very last metal ion complexes with the EDTA- when the solution turns completely blue. Because the quantitative reaction in this experiment is a complexation reaction, the chemistry is much slower than in acid-base titrations. As you get closer to the endpoint you need to allow the solution to sit for 20-30 seconds to be sure the reaction is complete. It is also necessary to a the EDTA drop wise while swirling the flask to mix thoroughly the reactants. This is a two-part experiment. The first art is the titration to determine the total concentration of ions in the unknown solution. This part of the experiment requires the use ofEBT as an indicator. The second part is the titration of only the calcium in the solution. In the second part, you will effectively remove all of the magnesium from the solution by adding hydroxide and titrate the calcium that is left. You will be using a different indicator for this part of the experiment (hydroxynaphthol blue, a solid) Experimental Procedure Preparing the solutions. pH 10 buffer: 28.5 mL concentrated ammonia (ammonium hydroxide) 3.5 gammonium chloride diluted to 50 mL with DI water. Store this in a tightly capped plastic bottle. EDTA solution: Accurately wei out-0.6 g ofNa2H2EDTA-2H2O (FM 372.24 g/mol). Dissolve the EDTA in a 500 mL volumetric flask that is full of DI water. Add 3 pellets of solid NaOH to your EDTA solution. Swirl the solution to dissolve the EDTA. (Ifit doesn't dissolve immediately, prepare your unknown and come back to this). Make sure your EDTA is completely dissolved and fill the flask to the mark with DI water and invert to mix Unknown solution: Accurately weigh out 0.25-0.35 g of your unknown directly into a clean, dry 250 mL Erlenmeyer flask. Use a minimum amount of-20% HCI (20 mL conc, HCI 80 mL DI water to dissolve the unknown solid. Your unknown is limestone and contains a mixture of calcium and magnesium carbonates it will bubble so when the acid is added be careful and add the acid slowly. You will know that the solid is completely dissolved w adding acid no longer results in the formation of bubbles (or the pH ofyour solution falls below 4).
Explanation / Answer
For calculating total moles from this complexomteric titration,
Average volume used = 37.73 ml = 0.03773L
Molarity of EDTA formed = 0.00322
but since, the data tells it to be 0.001624 M, we assume it to be same being used for the titration
Moles of EDTA = Molarity * Volume (L) = 0.001624*0.03773 = 0.0000612 moles or 6.12 *10-5 moles.
Since the molar ratio is 1:1, moles in unknown solution would be same i.e. 6.12 *10-5 moles
now, molarity of unknown solution containing total ions = moles/volume in L
the volume of unknown solution taken is = 10ml +20 ml water +3ml EBT =33ml =0.033L
molarity = 6.12 *10-5/0.033 = 1.85* 10-3M
For calculating moles of Ca2+ and Mg2+ from this complexomteric titration,
Average volume used = 27.82 ml = 0.02782L
Molarity of EDTA formed = 0.00322
but since, the data tells it to be 0.001624 M, we assume it to be same being used for the titration
Moles of EDTA = Molarity * Volume (L) = 0.001624*0.02782 = 0.00004518 moles or 4.518 *10-5 moles.
Since the molar ratio is 1:1, moles in unknown solution would be same i.e. 4.518 *10-5 moles
now, molarity of unknown solution containing Ca2+ = moles/volume in L
the volume of unknown solution taken is = 10ml +20 ml water +15ml =45ml =0.045L
concentration of Ca2+= 4.518 *10-5/0.045 = 1.004* 10-3M
concentration of Mg2+ = total concentration - concentration of Ca2+ = 0.84* 10-3M
EBT is not a suitable indicator for the second part of the experiment because EBT has the tendency to form red complex with all the metal ions. since to remove Mg ions we are adding NaOH, although the Mg would be removed, but the solution still will contain sodium ions. Using EBT will again complex with sodium ion as well and will result in the overestimation of the ions and thus will give wrong results.