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

So sometimes the question asks to subtract initial mole from final mole when do you do this?Capture.PNG
(edited 1 year ago)
Reply 1
anyone?
Original post by Mad Man
So sometimes the question asks to subtract initial mole from final mole when do you do this?Capture.PNG


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
Original post by BlueChicken
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).

But let me give you another example3.PNG4.PNG
Here you don't subtract anything.
Why?
(edited 1 year ago)
Reply 4
20220304_094534.jpg
20220304_094615.jpg
In jim Clarks book it gives an assumption
20220304_094639.jpg
When do you use this assumption and when do you not?
(edited 1 year ago)
Reply 5
Original post by BlueChicken
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).

How do you know if something is excess?
Reply 6
Can someone help?
Reply 7
Original post by Mad Man
20220304_094534.jpg
20220304_094615.jpg
In jim Clarks book it gives an assumption
20220304_094639.jpg
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 1 year ago)
Reply 9
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?
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
Original post by BlueChicken
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.

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.Screenshot_20220304-113534_Drive.jpg
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.Screenshot_20220304-113534_Drive.jpg

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
Original post by BlueChicken
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.

Righhht. I think I overcomplicated it all. Thanks 😊
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.

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