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### Topic: chemical equil. question...  (Read 4226 times)

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#### oxygen

• Guest ##### chemical equil. question...
« on: June 07, 2005, 01:55:07 PM »
Hello.

I am in need of help.

The question is: Calculate the volume of 2.00 M sodium bicarbonate to completely precipitate the lead(II) ion in 100mL of 0.0500M lead (II) nitrate solution buffered at pH 5.00.  Use 99.9% removal as the criterion for complete precipitation.

My problem is that I just don't know how to set up the chemical equations, mass-balance and charge balance.  I've tried almost everything from examples in the book and I just can't seem to solve this.  If you can help me figure out how to set up the equations, then I'm good from there on.  Thanks a bunch!

#### xiankai

• Chemist
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• Gender:  ##### Re:chemical equil. question...
« Reply #1 on: June 08, 2005, 05:13:55 AM »
i dont know about pH calculations, buffer solutions, but i know a few things about writting equations.

first we have NaHCO3 (aq) + Pb(NO3)2 (aq) --> 2NaNO3 (aq) + Pb(HCO3)2 (s)

the net ionic equation is:

HCO3- (aq) + Pb2+ (aq) --> Pb(HCO3)2(s)

unless there are other reactants formed (not really familar with bicarbonates), the ionic equation will be the key.

one learns best by teaching

#### hannibal

• Guest ##### Re:chemical equil. question...
« Reply #2 on: June 20, 2005, 06:08:36 AM »
first we need to neutralise the solution as sodium bicarbonate is a base

NaHCO3 + H+  --------> Na+   + CO2  + H2O
calculate this amount of sodium bicarbonate for neutralisation as with ph=5 we
have H+ = 10-5mol/litre therefore in 100 ml we have H+=10-6mol,so we need to first add 10-6 moles of sodium bicarbonate...now you should have the equillibrium constant(Ksp) for this equation -----
Pb(HCO3)2 -------->  Pb2+  +  2HCO3-  if this is given to be very small in that case you need to add 10-6  + 0.005 moles of sodium bicarbonate, accordingly calculate the volume of sodium bicarbonate needed....
but from the language of your question it appears that we need to take into account the Ksp value