Gen. Chem question on buffers

[snooper92]

Can someone help with this question?

Which of the following pairs of compounds could be mixed to produce a buffer solution?

A) CH3NH3I and HI
B) CH3NH2 and CH3NH3I
C) CH3NH3I and NH4I
D) HI and NH4I
E) CH3NH2 and NH3

The answer is "B." And the answer explanation states that this is a weak base and its conjugate base. I don't get it. Can you have a weak base and conjugate base?? With a buffer, aren't you looking for a weak acid and its conjugate base, or a weak base and its conjugate acid?

Also, how can you tell if a base is weak?

Can someone possibly explain how to solve -log (1.5 x 10^-6)? I don't understand why it's 6-log1.5. It's been a long time since I have done log calculations!

Thanks.

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[Hugh Mannity]

Yes it is weak acid and its CB or a weak base and its CA. Weak bases are all except the metal hydroxides I believe. Weak acids are all but the big 6. Ammonium is a weak acid as is ammonia (ammonia is also a weak base). An amine is a weak acid and a weak base, depending on the conditions. As long as you know not to mix any strong acid/bases, these should be manageable questions.

 

[harrygt]

Can someone help with this question?

Which of the following pairs of compounds could be mixed to produce a buffer solution?

A) CH3NH3I and HI
B) CH3NH2 and CH3NH3I
C) CH3NH3I and NH4I
D) HI and NH4I
E) CH3NH2 and NH3

The answer is "B." And the answer explanation states that this is a weak base and its conjugate base. I don't get it. Can you have a weak base and conjugate base?? With a buffer, aren't you looking for a weak acid and its conjugate base, or a weak base and its conjugate acid?

Also, how can you tell if a base is weak?

Can someone possibly explain how to solve -log (1.5 x 10^-6)? I don't understand why it's 6-log1.5. It's been a long time since I have done log calculations!

Thanks.

That is definitely a typo in the solution. You are rigth in the definition of buffer. You have strong acids in A and D, so those choices get eliminated. E has two weak bases, which makes no sense [eliminated]. remember, we are looking for weak base and its conj acid, or a weak acids and its conj base.
B and C are left. In C, we have to salts that will hydrolize in water to give us the conjugate acid of two weak bases. That is good, but where is the weak base ? [eliminated, as long as we don't have the weak bases in the mixture by themselves].
B is the perfect answer. CH2NH3 is a weak base. CH3NH3I will hydrolize to CH3NH3+ [the conjugate acid of that weak base] and I- in solution.
How do we know that CH3NH3 is a weak base? you have to know that NH3 is a weak base, and also all organic amines are considered to be weak bases too.
The stronge bases which you have to know are KOH, NaOH, Ba(OH)2, Ca(OH)2 and that is pretty much it.


For the second part:
Putt the minus away first till we get done with the log.
there is a formula: Log(x*y) = logx + logy
so we have: log 1.5 + log 10^-6 = log 1.5 + (-6) = log1.5 - 6
rembemer we still have a minue sign before this expression, so we will get 6 - log 1.5.
DONE!

 

[osimsDDS]

May I quickly add, the salt of either the weak base or weak acid is always coupled to a an anion or cation depending on if its a base or acid...thats why in the answers you see I-

Ex/ Acetic acid CH3COOH (acid) and its conjugate base CH3COO-Na+ (cation)

NH3 (ammonia is a weak base) and its conjugate acid NH4+Cl-

 

[snooper92]

Thanks so much for taking the time to write a detailed response! I really appreciate your help.

I want to make sure I understand what you are saying.... so, you said:

B and C are left. In C, we have to salts that will hydrolize in water to give us the conjugate acid of two weak bases. That is good, but where is the weak base ? [eliminated, as long as we don't have the weak bases in the mixture by themselves].
B is the perfect answer. CH2NH3 is a weak base. CH3NH3I will hydrolize to CH3NH3+ [the conjugate acid of that weak base] and I- in solution.

So, for "C," you are saying that in water, the result would be CH3NH3IH and NH4IH, (conjugate acids)?

And for "B," CH3NH3I is the salt, right? How does it become CH3NH3+?

For buffers, are you also looking for weak bases and their salts or weak acids and their salts?

Thanks again.

 

[osimsDDS]

Thanks so much for taking the time to write a detailed response! I really appreciate your help.

I want to make sure I understand what you are saying.... so, you said:

B and C are left. In C, we have to salts that will hydrolize in water to give us the conjugate acid of two weak bases. That is good, but where is the weak base ? [eliminated, as long as we don't have the weak bases in the mixture by themselves].
B is the perfect answer. CH2NH3 is a weak base. CH3NH3I will hydrolize to CH3NH3+ [the conjugate acid of that weak base] and I- in solution.

So, for "C," you are saying that in water, the result would be CH3NH3IH and NH4IH, (conjugate acids)?

And for "B," CH3NH3I is the salt, right? How does it become CH3NH3+?

For buffers, are you also looking for weak bases and their salts or weak acids and their salts?

Thanks again.

You are looking for either or, a buffer can either be a weak base and its conjugate acid or a weak acid and its conjugate base...in either case for B
you have CH3NH2 and CH3NH3I so the weak base would be the CH3NH2 because ammonia is a weak base and since bases Accept PROTONS the conjugate acid would be that where the weak base added a proton to it, salt or conjugate acid would be CH3NH3+I- (I- is an anion coupled to it, you can tell this is a salt because thats what it does most of the time)

 

[harrygt]

Thanks so much for taking the time to write a detailed response! I really appreciate your help.

I want to make sure I understand what you are saying.... so, you said:

B and C are left. In C, we have to salts that will hydrolize in water to give us the conjugate acid of two weak bases. That is good, but where is the weak base ? [eliminated, as long as we don't have the weak bases in the mixture by themselves].
B is the perfect answer. CH2NH3 is a weak base. CH3NH3I will hydrolize to CH3NH3+ [the conjugate acid of that weak base] and I- in solution.

So, for "C," you are saying that in water, the result would be CH3NH3IH and NH4IH, (conjugate acids)?

And for "B," CH3NH3I is the salt, right? How does it become CH3NH3+?

For buffers, are you also looking for weak bases and their salts or weak acids and their salts?

Thanks again.

For C, in water you will have CH3NH3+ and I - and also NH4+ and I-. CH3NH3+ and NH4+ are conjugate acids [strong] of weak bases. If we had a weak base like NH3 or CH3NH3, this would become a buffer. But all we have is the conjugate acids of the week bases. The ' I ' is just there to show the ionic compound as Osim said. It's like saying NH3, NH4Cl buffer system instead of NH3, NH4+ . As you know, pos and neg ions are always next to each other in solid form, and that is why they put a cation like 'I' next to the conjugate base or acid.

FOr B, yes, CH3NH3I is that salt. It is an ionic compound, made of CH3NH3+ and I-. An ionic compound simply dissociates to its ions in water.