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nojha1 · Nov 13, 2009, 10:48 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?

here,
I try to start the problem by writing following equation:
Pb2+(aq) + EDTA2-(aq) <-> PbEDTA(aq)
but I have no idea what to do next... help!

Perito · Nov 13, 2009, 11:24 AM

Originally Posted by nojha1

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?

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

$Pb^{2+} + EDTA^{2-} \rightarrow PbEDTA$

With this equation, you know that one mole of EDTA reacts with one mole of Pb(2+). You were told that the titration has an "end point". This implies that the amount of EDTA used was exactly the amount required to react with the Pb(2+) -- no more, no less.

1. You know that you had a 10 mL sample containing Pb(2+) and you're trying to determine how much lead is in it. The volume is so that you can figure out a concentration after you determine the amount of lead in the sample.

2. The EDTA concentration is given as $8.193 \times 10^{-4}\, \frac {moles}{liter}$

The units (moles/liter) or M (molar) is important.

3. You used 35.47 mL (0.03547 Liters) of solution.

First, calculate the number of moles of EDTA you used:

$0.03547\,\cancel {Liters} \, \times \, \frac {8.193 \times 10^{-4}\,moles}{\cancel {Liter}} =2.906 \times 10^{-5} \, moles \, (of\,EDTA)$

Note how the "Liters" cancels -- just like arithmetic. Since one mole of EDTA reacts with one mole of Pb(2+), this is also the number of moles of Pb in 10.0 mL of the sample.

Concentration is expressed in moles/liter. Can you now figure out the concentration of lead in the sample?

nojha1 · Nov 14, 2009, 10:05 AM

Originally Posted by Perito

$Pb^{2+} + EDTA^{2-} \rightarrow PbEDTA$

With this equation, you know that one mole of EDTA reacts with one mole of Pb(2+). You were told that the titration has an "end point". This implies that the amount of EDTA used was exactly the amount required to react with the Pb(2+) -- no more, no less.

1. You know that you had a 10 mL sample containing Pb(2+) and you're trying to determine how much lead is in it. The volume is so that you can figure out a concentration after you determine the amount of lead in the sample.

2. The EDTA concentration is given as $8.193 \times 10^{-4}\, \frac {moles}{liter}$

The units (moles/liter) or M (molar) is important.

3. You used 35.47 mL (0.03547 Liters) of solution.

First, calculate the number of moles of EDTA you used:

$0.03547\,\cancel {Liters} \, \times \, \frac {8.193 \times 10^{-4}\,moles}{\cancel {Liter}} =2.906 \times 10^{-5} \, moles \, (of\,EDTA)$

Note how the "Liters" cancels -- just like arithmetic. Since one mole of EDTA reacts with one mole of Pb(2+), this is also the number of moles of Pb in 10.0 mL of the sample.

Concentration is expressed in moles/liter. Can you now figure out the concentration of lead in the sample?

Thank You,!