# Buffer solutions

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#1
n(A–) = 9.25 × 10–3 (mol)
n(HA) = 0.0165 - 9.25 × 10–3 = 7.25 × 10–3 (mol)

This is what the mark-scheme says. 0.0165 is calculated from volume x concentration of the acid.

I put that HA = 0.0165 mol, I don't understand why you have to take away the moles of A- from this value to get the moles of HA?
0
1 year ago
#2
A buffer solution should have roughly equal amounts of moles/concentrations. So the moles of HA is in excess (because it has more moles than A-) So that’s why you subtract the moles with each other. Then you just carry on with the normal calculation. If that makes sense
1
1 year ago
#3
This method of preparing a buffer solution is via partially neutralising a weak acid.

You should know there are three main ways of preparing a buffer solution (adding a solid salt to a weak acid, adding a salt solution to a weak acid and adding an alkali to a weak acid to partially neutralise it to form a salt - you should have come across this before in pH curves where you have that "buffer region" before the end point - this is when a buffer solution was being formed"

Anyways back to the question:

I hope you are familiar with weak-acid and strong base neutralisation.

Find out moles of the base being added
Find out moles of the acid originally

Find out which moles is in excess
Here the acid moles is in excess so you have to use Ka to find pH (if moles of the base was in excess we'd have to use Kw to find pH)

Here you modify your Ka expression.

The A- now represents moles coming in from the base - as all these moles of base neutralised with the acid to form the salt.

The HA now represents excess moles of HA acid left over after the partial neutralisation took place.

I hope you can now see why we subtract the moles? we're just finding which is in excess - this is no different from a weak acid strong base type of calculation.

Now find conc. via c = n /v and plug it into Ka to find whatever the question was asking you to find.

Hope this helped !
Last edited by wendychan; 1 year ago
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