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Chemistry Forums for Students => High School Chemistry Forum => Topic started by: tubstar98 on December 05, 2017, 04:25:42 PM

Title: Ka pKa-calculations
Post by: tubstar98 on December 05, 2017, 04:25:42 PM
Hi everyone,
i have been given a assignment saying,

"Now that you know the concentrations of NH4, NH3 and H+, calculate the Ka and pKa of them."

I have 50 ml which consists of 10 ml 1m NaOh, 40 ml of water and 1 gram of NH4Cl.
pH is 9.5
my question is, do i use the formula "pH=pKa-log([NH_3 ]/[NH_4 ] )"?

My other question is, after i have mixed my 1 gram of CH4Cl in the 50ml, the NH4 turns into NH3 right? so when the reaction has occured, do i have anymore NH4?

sorry for my messy post.
Hope somebody can help, kind regards Tobias
Title: Re: Ka pKa-calculations
Post by: Borek on December 05, 2017, 04:28:50 PM
Now that you know the concentrations of NH4, NH3 and H+

Write formula for Ka.
Title: Re: Ka pKa-calculations
Post by: tubstar98 on December 05, 2017, 04:41:10 PM
So

Ka=[H3O+][NH3]/NH4? :/
Title: Re: Ka pKa-calculations
Post by: Borek on December 05, 2017, 04:45:09 PM
Which of these do you know?
Title: Re: Ka pKa-calculations
Post by: tubstar98 on December 05, 2017, 04:47:22 PM
I know the concentration of NH3 and H+ in the 50 ml liquid.

I only know the NH4 in grams.
Title: Re: Ka pKa-calculations
Post by: Borek on December 05, 2017, 06:01:48 PM
Mass and volume, sounds like concentration to me. Besides, you stated in your opening post, that

Now that you know the concentrations of NH4, NH3 and H+


so I see some contradiction here.
Title: Re: Ka pKa-calculations
Post by: tubstar98 on December 06, 2017, 02:34:49 AM
I see that.

I know how NH4 is in 1 gram NH4Cl, i do not know how much NH4 is in the 50ml. I thought all the NH4 molecules, when in aq, turned into NH3 molecules, so i thought the NH4 concentration in the 50ml was equal to zero. Is that a wrong assumption?
Title: Re: Ka pKa-calculations
Post by: Borek on December 06, 2017, 02:50:09 AM
There is no such thing as NH4 molecule. There is NH4+, ammonium cation, which is part of many salts.

Please elaborate on what the whole problem is, apparently it is either poorly defined or you are misunderstanding it.
Title: Re: Ka pKa-calculations
Post by: tubstar98 on December 06, 2017, 03:35:33 AM
we had an experiment in school, we took 40 ml of water, 10 ml of 1M NaOH and mixed them together along with 1,0005grams of NH4Cl. We then measured pH to 9.5.
i calculate the [H+]
〖10〗^(-9.5)=3,162278·〖10〗^(-10)

i write the reaction:
NH_4Cl+NaOH→NaCl+NH_3+H_2 O


i started calculating how many moles of NH4 was in 1,0005grams of NH4Cl.

0,01870391 mol NH_4 Cl in 1.0005gram
Then for OH-
0,004252213 mol OH in 10ml 1M NaOH

Then how much NH3 was after the reacton.
0,005956131 mol NH_3 after the reacton.

then i calculated the concentration of NH_3
0,1191226mol/liter.

My problem is, the assignment states that i have to calculate the concentration of NH4, are there any NH4 after the reaction occurs?
Same with OH-.

And
How do i calculate Ka and pKa?
Title: Re: Ka pKa-calculations
Post by: Borek on December 06, 2017, 03:39:45 AM
Just assume the reaction went to completion and follow the stoichiometry.

What is the limiting reagent?
Title: Re: Ka pKa-calculations
Post by: tubstar98 on December 06, 2017, 03:50:53 AM
Can just assume it goes to completion? or should i be able to see that?

So if the reaction goes to completion, then after the reaction has occured, there is no NH4 or OH  right?

The limiting reagent is OH?
Title: Re: Ka pKa-calculations
Post by: Borek on December 06, 2017, 06:17:33 AM
As with every assumption, you can check if it was valid after finishing calculations. (But in general, in most buffer calculations where we mix a weak acid/base with a strong one assumption that neutralization went to completion holds quite well).

So if the reaction goes to completion, then after the reaction has occured, there is no NH4 or OH  right?

No, you are misunderstanding "to completion". It doesn't mean "till there are no reagents", it means "till the limiting reagent is consumed".
Title: Re: Ka pKa-calculations
Post by: tubstar98 on December 06, 2017, 07:56:20 AM
Okay. i dont understand what to do then.
Title: Re: Ka pKa-calculations
Post by: Borek on December 06, 2017, 08:00:46 AM
How much NH4+ will be left after all NaOH reacted?
Title: Re: Ka pKa-calculations
Post by: tubstar98 on December 06, 2017, 10:11:04 AM
There will be the original amount of NH4 minus the amount NH4 which have given away an H atom to OH- to make H2O. the former posted calculations was wrong.

so NH4 left in the liquid will be: 0,01870391mol NH_4    minus   0,01mol NH_4==0,00870391 mol NH_4 .

so how much OH- is now in the liquid? have all of this reacted?
Title: Re: Ka pKa-calculations
Post by: tubstar98 on December 06, 2017, 10:12:05 AM
And there will be no NH3 before the reaction occurs?
Title: Re: Ka pKa-calculations
Post by: Borek on December 06, 2017, 11:10:00 AM
There will be the original amount of NH4 minus the amount NH4 which have given away an H atom to OH- to make H2O. the former posted calculations was wrong.

Yes.

Quote
so how much OH- is now in the liquid? have all of this reacted?

You can't ever use all OH- from the water, as because of water autodissociation there will be always some equilibrium concentration present. But for stoichiometry you can assume all of it reacted (and equilibrium concentration can be easily calculated from known pH).

And there will be no NH3 before the reaction occurs?

Small equilibrium amounts, safe to ignore.
Title: Re: Ka pKa-calculations
Post by: tubstar98 on December 06, 2017, 11:40:22 AM
I have now calculated my pKa value. It is 0,2 % deviant(spelling?) from what the assignment want me to compare it to. I have also calculated the [OH-] using the [H+].

I think i have understood what this assignment is about and how to calculate the problems,  it has a lot to do with the help you have given me, thank you very much for your time and effort in helping me understand. It is really appreciated!