Buffers (Gen Chem) Question

[humarathumara]

So as I understand it, the buffer system works by having an equal molar concentration of weak acid and conjugate weak base.

Example 1: In the Berkeley Review book they prove this by giving an example of a system where you have 999 parts HA and 1 part A-. Then the conc. of A-/HA = 1/999, then when you add a strong base, OH to the mix, you then have 2/998 conc. That essentially doubles the ratio, so since Ka = [H+] * [A-]/[HA], thus if the ratio of [A-]/[HA] doubles, then the H+ concentration must be halved to give a constant Ka value (ka is always constant with cons. temp).

Example 2: They then compare this to a system where you 500 parts HA to 500 parts A-. Then if you add OH- then the ratio A-/HA doesnt change much, thus H+ doesn't change much.

My question is this: How can you say in the first example that the ratio is 1 to 999, because reactants and products always come to equilibrium So even if you started with such an uneven ratio, it would eventually equal out to some value. That would be close to the same value that the second system with 500 parts of each would equal out to. Therefore, in both examples 1 and 2 the parts would equal out to some amount and then wouldn't that throw off all the calculations????

Thanks so much, buffers systems have always confused me.

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[tco]

In the first example, you're assuming that you're in a solution that would lead to an equivalent amount of products and reactants at equilibrium.

What if you're in an acidic solution where most of your bases are protonated?

 

[humarathumara]

I understand that, but then that would be the same case for example 2. So, if you had a solution where you have 999 parts HA to 1 part A-, and that was the equilibrium level. Then that would be true for example 2 as well. Meaning that when you put 500 of HA and 500 A-, it would equilibrate to something close to 999 to 1 again. You are not using a different HA between examples 1 and 2, so they should show the same equilibrium no matter what the starting conc.'s are.

 

[tco]

Okay, lets use real examples then.

What if I have an aqueous solution where I add 1 moles of HNO3?

What if I have an aqueous solution where I add 0.5 moles of HNO3 and 0.5 moles NaNO3?

Would the concentrations of free hydronium be the same?

It doesn't matter what the volumes are...Just assume one liter if you want...

You're correct in saying that the Ka will be the same for constant temperature. The amount that disassociates is dependent on the starting concentrations, however.

Now, what if we add another 0.5 moles of an acid to each? Which would have the larger change in pH?

pH=pKa - log ( [A-]\[AH] )

 

[humarathumara]

ok..... so you asked only questions....could i get the answers to those lol

again, when you used real examples u just supplemented what i said about the 999 to 1 parts and 500 to 500 parts with real nmbers. But that doesn't allow me to understsand the concept any clearer. Why is it that the ratio of 999 to 1 doesnt equilibrate to lets say, 900 to 100, and that the ratio of 500 to 500 doesnt ALSO equilibrate to 900 to 100? Why does this not happen? Thanks again

 

[humarathumara]

any answers would be deeply appreciated.