Chemistry Forum

Consider ortho-fluoro aniline (o-FC₆H₄NH₂).

a.Give an electron-dot formula for this compound and indicate with an arrow which atoms can accept a proton. Circle the atom that actually does accept a proton from water.

b.Determine whether o-FC₆H₄NH₂ is a stronger or weaker base than aniline itself (C₆H₅NH₂) and explain why this is the case.

c.Calculate the pH of a 0.1 M solution of o-FC₆H₄NH₂.

d.Draw the structural formula of the conjugate acid of o-FC₆H₄NH₂.

e.Calculate the pH of a 0.010 M solution of the conjugate acid of o-FC₆H₄NH₂.

What are answers to all these questions? How to answer all these questions? Let me answer all these questions one by one.

Answer to a.
Chemical structure of ortho-fluoro aniline is as follows:

Nitrogen atom can accept proton from water.

Answer to b.

o-FC₆H₄NH₂ is stronger base than C₆H₅NH₂ because F can act as base (electron pairs donor)

Both are wrong . What you draw is para fluor aniline not the ortho one.
And anilin pka 4,6 is stronger base as the fluoraniline pka 3,2. Reason strong - I effect of flourine, M - effect is miner.

Are Yoy talking about mesomeric effect and inductive effect of Fluorine?

Is this correct chemical structure drawing for ortho-fluoro aniline?

Now it is correct. Yes i talk about inductive and mesomeric effect.

Fluorine atom will accept proton from water in ortho-fluoro aniline.

You said pKa of the o-fluoro aniline is 3.2, so its pH is 3.2 as well.

Now, how to answer part d and e?

Wrong. The only atom which accept proton will be nitrogen to form the ammonium.
The calculation is similar the first exercise with the trimethylamin.

Let me answer part e.
Conjugate acid of o-FC₆H₄NH₂ is o-FC₆H₄NH₃
Reaction equation is FC₆H₄NH₂ + H₂O → FC₆H₄NH₃⁺ + OH⁻

So, 1.6E-11=x*x/0.01M → x= 4.0e-7, Hence pOH=-log(4.0e-7)= 6.4. so, pH=14.0-6.4=7.6 Is this computaion correct?

Where does the 0,00625 coming from. pH should be alkaline because its a amin

I edited the answer. Please check it out.

I dont know what you doing, where this numbers come from. pKa is 3.2 already

If pKa is 3.204, then ka= 0.000625, this means 0.000625* 0.01= x * x , → x= 0.0025 Now is this hydroxide ion concentration? If it is so, then poH=-log(0.0025)=2.602, this means pH= 14.000 -2.602=11.398 Is this computation correct?

Looks better now

How to use pH solver available on webqc.org to answer this question?

The given pKs is 3.2 for FC₆H₄NH₃+ , for the solver has to be used the corresponding pKb value of 10.8.
pH about 8.1