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when is it not a buffer solution?

Basically I have a question in which there are two solutions, A and B. A is a buffer because there is 25cm3 of sodium hydroxide 0.010 mol/dm3 and 50cm3 of 0.010 mol/dm3 of ethanoic acid. This apparently means it is acidic.

Howver B is not a buffer and apparently this is because it is neutral. The only difference is that it has 25cm3 of sodium hydroxide 0.020mol/dm3.

Why does this make it neutral? Because the concentrations are not the same? But why would it be neutral when I add more alkali?

:confused:

Reply 1

Wenzel

If you double the concentration, you'll double the amount of OH- ions (NaOH). So basically, doubling the concentration is the same as doubling the volume here. This means there is enough OH- ions to neutralise the acid in B (since there is equal moles of the ethanoic acid and sodium hydroxide and it is a 1:1 ratio; NaOH + CH₃COOH ----> CH₃COONa + H₂O).

In A, half of the H+ ions will be able to react, but half will still remain (since NaOH is half the volume but same concentration), leaving the system in equilibrium and slightly acidic, with approximately equal amounts of the salt and the acid i.e. a buffer solution. If you need an explanation on how buffer solutions work, feel free to ask.

Reply 2

-Kav-

This all seems like a long time ago, to the extent that I'm no longer sure what a buffer is, but if the question boils down to why B is neutral, it's because the overall concentrations of acid and alkali in the solution are the same. Since the neutralise in a 1:1 ratio, equal concentrations should result in a neutral solution.

Reply 3

hunibuni

Except that buffers act to minimise the pH change in an acid/base solution, I don't really get how they work tbh so an explanation would be great! Well...I mean, I get, for example, that with ammonium hydroxide, you would have an excess of the reactant and of the NH4+ ion so when you add OH- it can react with the NH4+ without there being a significant effect on pH and when you add OH- it reacts with the H+ and is removed as water and then the excess of the reactant means more can dissolve to replace the lost ions. But I am still a bit confused...and I don't quite get why the other half of the H+ ions cannot react - because you only effectively have 0.5 of the required OH- and so only 0.4 can be 'paired up'?

Also, how do I know it's a 1:1 ratio?

Thank you so much for your help, both of you! I'm sorry to be so annoying with all these questions!

Reply 4

Wenzel

What you've explained is more or less all there is too it. :p:

Using your example of ammonia, if you add small amounts of acid, the H⁺ ions will be removed by NH₃ and OH^-; since it's in equillibrium some H⁺ ions will remain, and the pH will still decrease by a very, very tiny amount (buffers oppose change in pH, don't stop it entirely). If you add a small amount of alkali NH₄⁺ will react with the OH^- ions producing NH₃ and H₂O; again though some OH^- will always remain.

I'll use my equation from earlier to explain the ratio thing:

NaOH + CH₃COOH ----> CH₃COONa + H₂O

Here you can see 1 mole of NaOH reacts with 1 mole of CH₃COOH, hence it is a 1:1. Now look at this equation:

2HCl + Na₂CO₃ ----> NaCl + CO₂ + H₂O

Now we need 2 moles of HCl for every 1 mole of Na₂CO₃ that reacts. So the molar ratio would be 2:1.