Buffer problem from Destroyer

[datdentist]

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Which of the following will not form a buffered solution?
100ml of .1M Na2CO3 and 50ml of .1M HBr
100ml of .1M NaHCO3 and 25ml of .2M HCl
50ml of .2M Na2CO3 and 5ml of 1M HCl
10ml of .25M NH3 and 20ml of .25M HCl
20ml of .25M NH4Cl and 10ml of .25KOH

Not really understanding the answer in the book. Would appreciate a thorough explanation, thanks!

 

[FeralisExtremum]

Buffer requires a 1:1 ratio of weak acid/base to weak conjugate base/acid or a 2:1 ratio of weak acid/base to strong conjugate base/acid. You should try to understand why logically: weak base and weak acid will cancel each other out if there are equal moles of both (which is an ideal buffer). With a 2:1 ratio of weak to strong, all of the strong reacts with half of the weak and we get a buffer leftover (example: if you have 10M of F- and add 5M HCl, after they react you get 5M of F- left, and formed 5M of HF - that's a 1:1 ratio of weak base to weak conjugate acid, so we inadvertently formed a buffer).

So what you need to do here is use the molarity formula to calculate how many moles of each acid/base you have, then make sure they are in the appropriate 2:1 ratio (you'll notice every single one of these is a weak:strong mixture, so look for a 2:1 ratio of moles of weak to moles of strong). I'll do the first one for you:

100ml of .1M Na2CO3 and 50ml of .1M HBr

molarity = moles/Volume, so for Na2CO3: .1M = (x moles)/.1L. Solving for x gives us .01 moles of Na2CO3.
For HBr: .1 M = (x moles)/(.050L). Solving for x gives us .005 moles of HBr.

Our mole ratio of weak base:strong acid here is 2:1, so this checks out as a buffered solution.

 

[DAT Destroyer]

Which of the following will not form a buffered solution?
100ml of .1M Na2CO3 and 50ml of .1M HBr
100ml of .1M NaHCO3 and 25ml of .2M HCl
50ml of .2M Na2CO3 and 5ml of 1M HCl
10ml of .25M NH3 and 20ml of .25M HCl
20ml of .25M NH4Cl and 10ml of .25KOH

Not really understanding the answer in the book. Would appreciate a thorough explanation, thanks!

This is actually a pretty easy problem. The main concept is this..... The solution can NOT contain a STRONG acid or BASE in EXCESS ....or EQUAL. If this occurs, you no longer have a buffer.

Note how each component is calculated in the solution....and hopefully the result will be evident.

Hope this helps.

Dr. Jim Romano