EZ Chem Q, Buffers
Started by BiomajorPreDent
I can't be much of help here. But I've also heard that you can use a salt. A solution usually containing an acid and a base, or a salt.
A (Bronsted) Acid is a proton donor and its Conjugate Base is what remains of the acid once it donates the proton. In this case, CH3NH3I is the (Bronsted) Acid and CH3NH2 is the Conjugate Base because the Acid donated a proton (H+) to the Bronsted Base.. So in simpler words, CH3NH3I has one more 'H' than the CH3NH2.. and also remember that Buffers can't be made with either Strong Acid or Strong Base..
Hmm.. I'm sure there are others who can elaborate better, but I hope it helped!
Hmm.. I'm sure there are others who can elaborate better, but I hope it helped!
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lol, i saw this one on the kaplan subject test that i did recently 😀
Trick here is that a buffer could be a weak acid and its salt, or a weak base and its salt.
CH3NH2 is a weak base (and its a base since it can accept an H) and CH3NH3I is its salt. Just think of it as CH3NH3+ I- in solution.
A good example of a buffer would be what we have in our body. H2CO3 is a weak acid and it makes a buffer with something like NaHCO3. See here, HCO3- is the salt/conj base of H2CO3.
Trick here is that a buffer could be a weak acid and its salt, or a weak base and its salt.
CH3NH2 is a weak base (and its a base since it can accept an H) and CH3NH3I is its salt. Just think of it as CH3NH3+ I- in solution.
A good example of a buffer would be what we have in our body. H2CO3 is a weak acid and it makes a buffer with something like NaHCO3. See here, HCO3- is the salt/conj base of H2CO3.
Oh, ok cuz i was getting confused because theres an I in one and no I in the second...
