When do you subtract values when finding pH of buffer solutions?

Mad Man

So sometimes the question asks to subtract initial mole from final mole when do you do this?

(edited 2 years ago)

anyone?

Original post by Mad Man

So sometimes the question asks to subtract initial mole from final mole when do you do this?

The bit you have circled has initial (0.018) - used (0.0015) = final (0.0165). Am I missing something? You may be provided with different information at the start of the question, so that you need to determine, say, the initial amount, but it would just be rearranging that equation. If you subtracted initial from final, it would give you the amount used up in the reaction.

The point of the circled bit is to determine, after the reaction, how much acid is remaining (as it starts in excess).

Reply 3

Mad Man

Original post by BlueChicken

But let me give you another example

Here you don't subtract anything.
Why?

(edited 2 years ago)

Reply 4

Mad Man

In jim Clarks book it gives an assumption

When do you use this assumption and when do you not?

(edited 2 years ago)

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Reply 5

Mad Man

Original post by BlueChicken

How do you know if something is excess?

Can someone help?

Original post by Mad Man

In jim Clarks book it gives an assumption

When do you use this assumption and when do you not?

The assumption is made when you are determining the pH of a weak acid (see p. 196-197 of Jim Clark's book - also see chapter on Buffer solutions and what they are starting on p. 206).

When you make a buffer, if it is a weak acid/base and one of its salts, you've made an acidic/basic buffer and you use the formula in post #4.

However, you can also make an acidic buffer, by taking a weak acid and adding a small amount of alkali (i.e. what is shown in post #1). A reaction would take place, neutralising some of the acid and forming a salt. The resulting solution then contains a weak acid and its salt (an acidic buffer - see above). Here the acid was in excess, as you added a small amount of alkali. So, here, you need to do the calculation from post #1, as you need to work out how much weak acid and its salt you have remaining after the reaction, so that you can determine the pH.

It looks like you might have the CGP chemistry book - if so, see p.358 to 363.

Hope that helps.

(edited 2 years ago)

Reply 9

Mad Man

Original post by BlueChicken

The assumption is made when you are determining the pH of a weak acid (see p. 196-197 of Jim Clark's book - also see chapter on Buffer solutions and what they are starting on p. 206).

So when I am not finding the pH of a weak acid I do find something by subtraction?

Reply 10

BlueChicken

Original post by Mad Man

So when I am not finding the pH of a weak acid I do find something by subtraction?

See my post above. You’ll need to work out how much acid and salt you have after a reaction if you made a buffer by adding a small amount of alkali to a weak acid. If the buffer is made from a weak acid and its salt, then you know how much you have of each.

As Jim says, you need to learn why you are doing something, not just an equation, as the exam question might be worded differently. Revisit both Jim’s book and the CGP book.

Reply 11

Mad Man

Original post by BlueChicken

Okay I will but FTR, how do you know if the acid/base is strong without knowing the pH?
In the chemrevise website it says you subtract for strong bases.

Reply 12

BlueChicken

Original post by Mad Man

Okay I will but FTR, how do you know if the acid/base is strong without knowing the pH?
In the chemrevise website it says you subtract for strong bases.

I think you need to move away from the idea of "subtract". The above just says the same as I had said above. If you are making the buffer from an acid and a base, rather than an acid/base and one of their salts, then you need to work out what is left over after the reaction between the acid and base has occurred.

Reply 13

Mad Man

Original post by BlueChicken

Righhht. I think I overcomplicated it all. Thanks 😊

Reply 14

BlueChicken

Original post by Mad Man

Righhht. I think I overcomplicated it all. Thanks 😊

It's easily done! (especially when stressing for exams). Framing the problem correctly if key - once, you done that, the rest is pretty much rearranging or putting numbers into an equation. Good luck with it.