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/sci/ - Science & Math

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5001064 No.5001064 [Reply] [Original]

Any chemists around?

This seems extremely simple, still, I can't seem to be able to solve it.

I tried using:


<span class="math">pH = pK_a + log \frac{[A^{-}]}{[HA]}[/spoiler]
as <span class="math">pH=pK_a[/spoiler]
<span class="math">log \frac{[A^{-}]}{[HA]} = 0[/spoiler]
<span class="math">\frac{[A^{-}]}{[HA]} = 1[/spoiler]
<span class="math">[A^{-}] = [HA][/spoiler]


So I've written the equilibrium reaction: (which I had to google, how would I know it's not a base?)

<span class="math">CH_3COOH + H_2O \rightleftharpoons H_3O^+ + CH3COO^-[/spoiler]


So, I tried finding <span class="math">K_a[/spoiler]

<span class="math">K_a = \frac{[CH_3COO^-][H_3O^+]}{[CH_3COOH]}[/spoiler]
And as I discovered <span class="math">[A^{-}]} = {[HA][/spoiler]


<span class="math">K_a = \frac{[x][H_3O^+]}{[x]}[/spoiler]
<span class="math">K_a = [H^+][/spoiler]
<span class="math">pK_a = pH[/spoiler]

See? I've been redundant during the entire time and couldn't find an answer...

Could someone, please, point the simplest way of solving this? Thank you.

>> No.5001126 [DELETED] 

bump

>> No.5001178

Okay, forget all garbage I've written.

Can someone help me solving it?

>> No.5001208 [DELETED] 

bump

>> No.5001240

You still need to figure how much NaOH you need to add...
writing CH3COOH = H+ + CH3COO- is a good idea.
You understand that the solution is too acid to be used as a buffer solution, so you want to get a higher pH, and to do that you can indeed add HO-.
Now write the reaction and all the related stuff!

>> No.5001244

>>5001240

Isn't CH3COOH a weak acid and CH3COO- its conjugated base?

Doesn't "weak acid+its conjugated base = buffer"?
I don't get why it's needed to add NaOH

>> No.5001267

>>5001244
usually, a buffer solution is made such as pH=pKa (so you have to use a weak acid anyway).
But you still need to have the good proportions.

>> No.5001271

>>5001267

But, I mean, wouldn't CH3COOH dissociate in CH3COO- and already make a buffer?

Why s it needed to add NaOH?

Could you show me how to solve the exercise in the simplest way possible?

Thank you

>> No.5001304 [DELETED] 

.

>> No.5001311

>>5001271
ok, as CH3COOH (let's call it ROH) is a weak acid, and you know the pH (pH=4), you can determine [RO-]/[ROH] by using the expression of pH in terms of pKa.
But you also know the initial concentration in ROH ( [ROH]=2.0 mol/L)
you can therefore deduce [RO-], right?

(do they give you the pKa? it's around 4.7 if I remember well)

whatever:
you know [RO-] and [ROH], and those are different since initially pH != pKa, but you want them to be equal.

This is why you need to adjust the concentrations.
By adding NaOH, HO- will mainly react with ROH to give RO-;
I guess you can go on your own from now on?
I'm staying for a while and I'll check later if you still have any questions.

>> No.5001325

>>5001311

<span class="math"> pH =pK_a \frac{[RO-]}{[ROH]} [/spoiler]
<span class="math"> 4 = 4.18 \frac{[RO^-]}{2} [/spoiler]
<span class="math"> -0.36 = [RO^-] [/spoiler]

Okay, what is wrong?
I tried exactly what you said, but there cannot be negative concentration...

>> No.5001362

>>5001325
what?

pH=pKa + log([RO-]/[ROH])
so log([RO-]/[ROH])=pH-pKa
=> [RO-]/[ROH]= 10^(pH-pKa)
=>[RO-]=[ROH]*10^(pH-pKa)!

>> No.5001389

>>5001362

I forgot the + sign. Anyway:


pH=pKa + log([RO-]/[ROH])
so log([RO-]/[ROH])=pH-pKa
as pH = 4 , pKa = 4.18 and [ROH] = 2
=> log[RO-]/2= 4 - 4.18
=>log[RO-]= -0.36
=>[RO-] = 0.4

Oh, yeah, correct result now.

So, with I continue as:

[ROH] = 2
[RO-] = 0.4

They need to be equal.
So, I need to add 0.8 of OH- ?
Which is also 0.8 of NaOH

Is this the answer?

>How many moles of NaOH must be added?
0.8 moled

>> No.5001400

>>5001389
yep, I think it is the right answer!

>> No.5001403

>>5001400

Thank you!

So, what I learned from this was:

"A weak acid dissociates into its conjugated base, but by itself, it's not a buffer. It needs to have a pH that's equal to its pKa"

Is that correct?

>> No.5001746

>>5001403
A weak acid CAN by itself act as a buffer. If a base is added to the solution, it will neutralize the weak acid, resulting in only a very small pH change. But without a conjugate base, it isn't a true buffer, because it is unable to resist pH change in both directions. A buffer need only consist of a weak acid/base and its conjugate base/acid. The concentrations don't have to be equal for it to be a buffer, but when they are, it forms the most effective buffer. At this point, pH = pKa. Usually, when you are trying to make a buffer with a specific pH, you will not end up with equal concentrations of acid and conjugate base.