[payata]

 

Hey Guys,

I just started training with exercises, pH and pKa value
I wanted to ask someone to help me understand, and to explain step by step how they got value of : 6.3 x 10-5

Here is a task:
The acid dissociation constant (Ka) for benzoic acid is 6.3 x 10-5. Find the pH of a 0.35 M solution of benzoic acid.

pKa = [H+][ C7H3O2-]/[HC7H3O2] = (x)(x)/0,35 = 6.3 x 10-5

Thank you in advance!

[Borek]

Not sure what you are asking, Ka is given, doesn't matter where they got it from.

Plus,

pKa = [H+][ C7H3O2-]/[HC7H3O2] = (x)(x)/0,35 = 6.3 x 10-5

This is wrong in two ways. First, it is not pKa, it is Ka on the left. Second - going to the place where x appears requires assuming x << 0.35, so the equal sign should be replaced by ≈ (approximately equal).

[AWK]

The solubility of benzoic acid at 25°C (298 K) is very small. The saturated solution at this temperature is below 0.03 M

[payata]

Oh, ok . I am a bit still confuse:

for example, teacher gave us to solve, find pH of 0.1M acetic acid.That's it, no more info. Therefore I was looking for pKs value which is 4, 75 . but no:

When I look to internet, and everywhere is given Ka value?, How can I know that I need to look up for ka value? how to know the diffrents where to you pka, ka. He is also using aberr. of pKs, is it the same as pKa and pKb 

If somebody somebody can explain it very easy I would be grateful 

[AWK]

pK_a=-log(K_a)

K_a=10^-pK_a

[payata]

So, to find Ka, I will use Ka=10-pKa, and to find to find pKa: pKa=-log(Ka) .. ?

And what about pKb, pKs?

In last example, why I have to use Ka value?

[Borek]

So, to find Ka, I will use Ka=10-pKa, and to find to find pKa: pKa=-log(Ka) .. ?

By definition.

And what about pKb, pKs?

pK_whatever always means -log(K_whatever).

In last example, why I have to use Ka value?

Because that's what directly describes the dissociation equilibrium.

[payata]

Clear! Thank you!