Analysis of Tap water (Hardness)

Was analyzed by a complexometric titraion by the following equations 1 & 2

1. Ca^2+ + Mg^2+ --> Ca^2+ + Mg(Indicator)

2. Mg(Indicator) + Ca^2+ + EDTA^4- --> CaEDTA^2- + MgEDTA^2- + free indicator

For the volume of titrant used (9.95 mL) calculate the:

a) Ca^2+ + Mg^2+ in sample (mol)
b)[Ca^2+] + [Mg^2+] of sample (M)

PLEASE HELP! I am very confused with this question and any guidance would be greatly appreciated!!!

I am confused by the question also. Is this a problem to be solved or is it a procedure that you have summarized with the two equations? I may be able to help if I know the procedure followed. The complexometric titration with EDTA is usually straight forward.

The equations 1 & 2 were given, and I believe we are supposed to make an equation for the reaction from this i.e. add them so that:

Ca^2+ + Mg^2+ + EDTA --> MgEDTS^2- + CaEDTA-

So by using the given volume of titrant (EDTA=9.95 mL) we could find the mol of Ca and Mg...i'm just not sure how to go about doing this (the stoichiometry is confusing me since the concentration they are looking for is Ca and Mg added...)

To solve this question, we need to use the information given and apply some stoichiometry and titration calculations.

Let's start by analyzing the given equations:

1. Ca^2+ + Mg^2+ --> Ca^2+ + Mg(Indicator)
This equation shows that calcium ions (Ca^2+) and magnesium ions (Mg^2+) react with each other. The reaction does not consume any of the calcium ions, only the magnesium ions are converted into a complex with the indicator. This equation does not affect our calculations directly.

2. Mg(Indicator) + Ca^2+ + EDTA^4- --> CaEDTA^2- + MgEDTA^2- + free indicator
This equation represents the actual complexometric titration, where the calcium ions (Ca^2+) and magnesium ions (Mg^2+) react with the EDTA^4- (ethylenediaminetetraacetic acid) to form complexes. The calcium ions are converted into CaEDTA^2-, and the magnesium ions are converted into MgEDTA^2-, while the excess EDTA^4- forms a complex with the indicator.

Now, let's move on to solving the problem:

a) To calculate the moles of Ca^2+ + Mg^2+ in the sample, we need to determine the concentration of EDTA^4- in the titrant solution.

Given:
Volume of titrant used = 9.95 mL

Since the volume of the titrant solution used is given, we can consider that the stoichiometric ratio is 1:1 between the titrant solution (EDTA^4-) and the Ca^2+ + Mg^2+ ions in the sample. This means that the number of moles of EDTA^4- used is equal to the number of moles of Ca^2+ + Mg^2+.

To calculate the moles of EDTA^4- used, we need the concentration of EDTA^4- in the titrant solution. This information is not provided in the question, so you will need to refer to the experimental procedure or any additional given data to obtain this concentration. Once you have the concentration of EDTA^4-, you can calculate the moles of Ca^2+ + Mg^2+ in the sample.

b) To calculate the [Ca^2+] + [Mg^2+] concentration in the sample, we need the volume and concentration of the sample.

Given:
Volume of titrant used = 9.95 mL
Assuming there is no dilution or additional components in the sample, we can assume that the volume of the sample is also 9.95 mL.

Now, if we assume there is no change in volume upon titration, the volume of the sample is the same as the volume of the titrant used. In this case, the total volume of the sample and titrant is 9.95 mL + 9.95 mL = 19.9 mL.

Using the moles of Ca^2+ + Mg^2+ calculated in part a), you can now find the molar concentration ([Ca^2+] + [Mg^2+]) of the sample by dividing the moles by the total volume in liters (19.9 mL = 19.9 x 10^-3 L).

Please note that if there are other factors or additional information given, you should refer to those as well to ensure accurate calculations.