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Topic: How could I make a buffer solution if I only had the weak base NH3 and HCl?  (Read 2168 times)

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Offline LackOfNames34

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How could I make a buffer solution if I only had the weak base NH3 and HCl?
« on: May 03, 2021, 10:50:06 AM »
I'm not sure how to answer this. Someone briefly explained that adding 1 mol of HCl and 2 mol of NH3 would lead to an NH3/NH4 buffer. But I don't really understand that. Any help?
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Offline Babcock_Hall

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Re: How could I make a buffer solution if I only had the weak base NH3 and HCl?
« Reply #1 on: May 03, 2021, 10:51:54 AM »
A good place to start is to write a balanced chemical equation.
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Offline LackOfNames34

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Re: How could I make a buffer solution if I only had the weak base NH3 and HCl?
« Reply #2 on: May 03, 2021, 11:23:04 AM »
Ok so a balanced equation would be NH3+HCl = NH4 + Cl
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Offline penta-d

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Re: How could I make a buffer solution if I only had the weak base NH3 and HCl?
« Reply #3 on: May 03, 2021, 12:29:02 PM »
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Offline LackOfNames34

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Re: How could I make a buffer solution if I only had the weak base NH3 and HCl?
« Reply #4 on: May 03, 2021, 12:58:36 PM »
I'm still a little bit lost. So we would react the NH3 and HCL to form NH4Cl. But how does this lead to NH3 +H2O = NH4 +OH?
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Offline Meter

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Re: How could I make a buffer solution if I only had the weak base NH3 and HCl?
« Reply #5 on: May 03, 2021, 01:40:49 PM »
Quote from: NoCreativeNames on May 03, 2021, 12:58:36 PM
I'm still a little bit lost. So we would react the NH3 and HCL to form NH4Cl. But how does this lead to NH3 +H2O = NH4 +OH?
1 mole of HCl reacts with 2 moles of NH3. By the end, all your HCl will be gone but you will have an equilibrium mixture of NH3 and it's conjugated acid, NH4+.
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Offline LackOfNames34

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Re: How could I make a buffer solution if I only had the weak base NH3 and HCl?
« Reply #6 on: May 03, 2021, 01:56:39 PM »
Ohh so by doing so, I would be reacting the HCl w NH3 to completion. And the equilibriums for the NH3 and the conjugate acid Nh4+ would be

NH3 + H2O <—> NH4+ + OH-

Or

NH4+ <—> NH3 + H+

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Offline Meter

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Re: How could I make a buffer solution if I only had the weak base NH3 and HCl?
« Reply #7 on: May 03, 2021, 02:39:34 PM »
Quote from: NoCreativeNames on May 03, 2021, 01:56:39 PM
Ohh so by doing so, I would be reacting the HCl w NH3 to completion. And the equilibriums for the NH3 and the conjugate acid Nh4+ would be

NH3 + H2O <—> NH4+ + OH-

Or

NH4+ <—> NH3 + H+
So you are adding a strong acid to an aqueous solution of ammonia. What exactly is happening?
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Offline LackOfNames34

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Re: How could I make a buffer solution if I only had the weak base NH3 and HCl?
« Reply #8 on: May 03, 2021, 02:44:01 PM »
Wouldn't the added acid be consumed in the buffer solution? Did I make an error?
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Offline Meter

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Re: How could I make a buffer solution if I only had the weak base NH3 and HCl?
« Reply #9 on: May 03, 2021, 02:47:10 PM »
Quote from: NoCreativeNames on May 03, 2021, 02:44:01 PM
Wouldn't the added acid be consumed in the buffer solution? Did I make an error?
The protons from the acid would react with ammonia to form NH4+, but after the reaction is over, an equilibrium will be reached. Your question is which equilibrium is reached, correct?
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Offline LackOfNames34

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Re: How could I make a buffer solution if I only had the weak base NH3 and HCl?
« Reply #10 on: May 03, 2021, 02:49:02 PM »
Yes, furthermore the pH would not have changed much if im not mistaken since NH4+/NH3 is a buffer solution
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Offline Borek

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Re: How could I make a buffer solution if I only had the weak base NH3 and HCl?
« Reply #11 on: May 03, 2021, 03:36:35 PM »
Quote from: NoCreativeNames on May 03, 2021, 02:44:01 PM
Wouldn't the added acid be consumed in the buffer solution?

Why would that matter?

What is the acid in your buffer solution? What is the base?
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Offline Babcock_Hall

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Re: How could I make a buffer solution if I only had the weak base NH3 and HCl?
« Reply #12 on: May 03, 2021, 04:07:40 PM »
Quote from: NoCreativeNames on May 03, 2021, 02:49:02 PM
Yes, furthermore the pH would not have changed much if im not mistaken since NH4+/NH3 is a buffer solution
This seems like the key point to me.
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