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nojha1 · Nov 20, 2009, 10:09 AM

A 10.00 mL sample of the water from the holding tank requires 35.47 mL of 8.193 x 10-4 M EDTA to reach the end point. (A) Based on this titration, what is the concentration of lead(II) ions in the holding tank? (B)Suppose a terrorist group sends a letter to the mayor claiming they have dumped 2 tons of lead(II) nitrate into a 500,000 gallons holding tank at a local reservoir. (Have the terrorists been “accurate” as to their claim?) AND, Is the concentration above or below the EPA acceptable level of 15 ppb?

here,
I try to start the problem by writing following equation:
Pb2+(aq) + EDTA2-(aq) <-> PbEDTA(aq)

One of the member with ID PERRITO helped me with second step as,

The number of moles of EDTA = 0.03547 liter X (8.193 x 10-4 moles/Liter)
= 8.193 x 10-4 moles of EDTA
Since, since one mole of EDTA reacts with 1 mole of pb(2+) so, the number of moles of pb in 10.00 ml is also same.

Then, I DID,
number of moles of pb(2+) in 10.00 ml solution = 8.193 x 10-4 moles of pb
So, Molarity or concentration of pb(2+) = (moles/lit)

=8.193 x 10-4 moles of pb/(10.00X10^-3Lit)
= 0.08193 M [pb(2+)]
IS that right? AND
CAN anyone please help me with part B!!

Unknown008 · Nov 21, 2009, 01:46 AM

Yes, the answer is correct.

Unfortunately, I'm not familiar with the english gallon.

You need to find the amount of moles in 2 tonnes of lead nitrate. Then, you'll be able to find the concentration of lead (II) in the 500 000 gallon of water. Then, convert it into 'ppb' and compared your value with 15 ppb.

You could have continued from your thread there:
https://www.askmehelpdesk.com/math-s...on-415664.html