[Fragment_Error]

Hello everyone. I joined here, because for some reason, of all the sciences, chemistry is the one I have most trouble with. Don't know why. I can understand math, physics and biology easily, but chemistry always gives me trouble. I'm 20 and studying Biotechnology and Forensic science at university and am having trouble with the chemistry subject. Just thought I'd give a little background before getting onto the questions.

Well on to the qustions.

Part 1:

I'm pretty sure I'm right on these, but I just need some confirmation. Please tell me if these equations are balanced for the reactions occuring:

HCl + NaOH -> NaCl + H₂O (ratio of HCl to NaOH is 1:1)
CH₃COOH + NaOH -> CH₃COONa + H₂O (ratio of CH₃COOH to NaOH is 1:1)
CH₃COOH + N_H4OH -> CH₃COONH₄ + H₂O (ratio of CH₃COOH to NH₄OH is 1:1)


Part 2:

Now the next qustion is where I had a graph and had to plot the equivalence point of HCl vs NaOH.

Alright, now the equivalence point on my experiment was a pH of 6 at about 26.6mL

Now to find the moles of NaCl formed I first find the atomic mass of NaCL, then multiply by Avogadro's number:

Atomic mass of NaCl: 22.99 + 35.45 = 58.44
Moles of NaCl formed: 58.44 x (6.022 x 10²³) = 3.42 x 10²⁵ Mol


Part 3:

The next part is the concentration of CH₃COONa at the equivalence point.

The formula is:

moles x 1000/total volume in mL (total volume in mL is the volume in the burette? So that's 25mL right?)

So the amount of moles in CH₃COONa is: 12.01 + (1.008 x 3) + 12.01 + 16 + 16 + 22.99 = 82.034

82.034 x (6.022 x 10²³) = 4.94 x 10²⁵ Mol

Now back to the formula:

(4.94 x 10²³) x 1000/25 = 1.976 x 10²⁷ mol/dm³


Part 4:

Now this part is the part that I'm not too sure of:

It has an equation:

pH = 1/2pK_w + 1/2pK_a + 1/2 log c

What on earth do I do here? What's p, k, w, a and c? What numbers do I plug in there?



I've tried to do this as much as I can, and am pretty sure I've got everything right, except for part 4, which I have no idea what to do there.

Any help would be greatly appreciated. Thank you for taking the time to read the thread anyway.

[Dan]

Part 1 is fine, but after that you're in a mess I'm afraid.

I'm assuming you have your base (NaOH or NH₄OH) in the burette. Is that right?

Now to find the moles of NaCl formed I first find the atomic mass of NaCL, then multiply by Avogadro's number:

NO. You know how much NaOH you added, and the concentration of NaOH solution. From this you can work out the number of moles of NaOH. From that you can work out the number of moles of NaCl produced.
You don't do anything with Avagadro's number.

The next part is the concentration of CH₃COONa at the equivalence point.

The formula is:

moles x 1000/total volume in mL (total volume in mL is the volume in the burette? So that's 25mL right?)

NO. You know how much NaOH you added, and the concentration of NaOH solution. From this you can work out the number of moles of NaOH. From that you can work out the number of moles of CH₃COONa produced. The total volume is the the total volume of the solution containing CH₃COONa. It is the volume of the acid you started with, PLUS the volume of base you added.

Now this part is the part that I'm not too sure of:

It has an equation:

pH = 1/2pK_w + 1/2pK_a + 1/2 log c

What on earth do I do here? What's p, k, w, a and c? What numbers do I plug in there?

pK = -logK

K_w is the equilibrium constant for the equilibrium 2H₂O <------> H₃O⁺ + OH^-
The value of this is usually taken as 10^-14 mol² L^-6

K_a is the equilibrium constant for the equilibrium, where AH is your acid,  AH <----> A^- + H⁺
You will have to work that out.

c is concentration of the acid (AH) used - which you know.

[Dan]

When you do calculations, think about the numbers you are getting, are they sensible? You should realise that 1 mol of NaCl has a mass of about 58.5g.
Is it sensible to say that you produced 3.42 x 10²⁵ mol of NaCl in your titration experiment? Do you remember having to hire an enormous convoy of dumper trucks to tidy away two thousand billion billion Tonnes of NaCl after the experiment?

[Fragment_Error]

Part 1 is fine, but after that you're in a mess I'm afraid.

I'm assuming you have your base (NaOH or NH₄OH) in the burette. Is that right?

Yes, that's right.

NO. You know how much NaOH you added, and the concentration of NaOH solution. From this you can work out the number of moles of NaOH. From that you can work out the number of moles of NaCl produced.
You don't do anything with Avagadro's number.

Ok. The concentration of the NaOH was 0.1M, the amount of NaOH I used at the equivalence point was 26.6mL. So does that mean I go 0.1/26.6 = 0.3 x 10^-3 Mol?

If so, how do I find out how much NaCl I make out of it?

NO. You know how much NaOH you added, and the concentration of NaOH solution. From this you can work out the number of moles of NaOH. From that you can work out the number of moles of CH₃COONa produced. The total volume is the the total volume of the solution containing CH₃COONa. It is the volume of the acid you started with, PLUS the volume of base you added.

How do I know how much CH₃COONa it produces? Seriously, I have no idea....

pK = -logK

K_w is the equilibrium constant for the equilibrium 2H₂O <------> H₃O⁺ + OH^-
The value of this is usually taken as 10^-14 mol² L^-6

K_a is the equilibrium constant for the equilibrium, where AH is your acid,  AH <----> A^- + H⁺
You will have to work that out.

c is concentration of the acid (AH) used - which you know.

Alright so:

K_w = 10^-14
K_a - how do I work out the equilibrium constant?
When you say (AH) what does it stand for?


Sorry. As I said eariler, I just really don't get chemistry.

When you do calculations, think about the numbers you are getting, are they sensible? You should realise that 1 mol of NaCl has a mass of about 58.5g.
Is it sensible to say that you produced 3.42 x 10²⁵ mol of NaCl in your titration experiment? Do you remember having to hire an enormous convoy of dumper trucks to tidy away two thousand billion billion Tonnes of NaCl after the experiment?

Haha, sorry. I tend to lose grasp and don't look at it to see if it makes any sense

[Dan]

Ok. The concentration of the NaOH was 0.1M, the amount of NaOH I used at the equivalence point was 26.6mL. So does that mean I go 0.1/26.6 = 0.3 x 10^-3 Mol?

No, you need to watch the units! your concentration, 0.1M means 0.1 mol L^-1, so to find the number of moles, you MULTIPLY the concentration by the volume IN LITRES. Compare the units on each side, 
mol L^-1 = mol / L,
so mol = mol L^-1 x L

in other words, number of moles = concentration x volume (in litres)

If so, how do I find out how much NaCl I make out of it?

Write a balanced equation. (Hint: how many moles of NaCl will 1 mol of NaOH generate?)

How do I know how much CH₃COONa it produces? Seriously, I have no idea....

Same method as above, look at your balanced equation.

Alright so:

K_w = 10^-14
K_a - how do I work out the acidity constant?
When you say (AH) what does it stand for?

K_a = [A^-][H⁺]/[AH]

example, for the equilibrium CH₃COOH <----> CH₃COO^-  + H⁺
AH = CH₃COOH
so A^- = CH₃COO^-

The square brackets mean 'concentration of'

You know [CH₃COOH]
You have worked out [CH₃COO^-] (it is the same as [CH₃COONa])
And if you look at the balanced equation I wrote above, you can see that the amount of CH₃COO^- produced is equal the amount of H⁺.

and therefore [CH₃COO^-]=[H⁺]

And therefore

K_a = [CH₃COONa]²/[CH₃COOH]

If you still can't get your head around this, I strongly suggest you discuss the concepts with your tutor. This is vey basic chemistry that you really should understand at degree level.

[Borek]

pH = 1/2pK_w + 1/2pK_a + 1/2 log c

Dan have already explained symbols used, so only some additional info.

HA (or AH) usually stands for any weak acid with dissociation reaction equation:

HA <-> H⁺ + A^-

It can be HCl (A^- = Cl^-, hardly weak), or CH₃COOH (A^- = CH₃COO^-) and so on.

K_a constants for different acids are given in the tables, you don't calculate them, you look them up Check pKa values link ine the left menu, or try my short list of pKa/pKb values.

Now, the formula you are given

pH = 1/2pKw + 1/2pKa + 1/2 log c

is completely wrong. For a weak acid (if some additional conditions are fulfilled) following formula works:

[H⁺ ] = sqrt(K_a*c)

which can be written as

pH = 1/2 pKa + 1/2 log c

Adding 1/2 pKw doesn't make sense. I suppose it is a modified version of equation

[H⁺ ] = 10^-7 + sqrt(K_a*c)

Which tries to overcome problem that may arise when you omit water dissociation when calculating pH of a very diluted acid - but does it wrong. Take a look at this lecture on weak acid pH calculation, once you wll understand what is going on try this pH cheat sheet.

Hm, Dan already posted something again... But I will add my $.02 anyway, I hate to delete so much typed text

[Dan]

pH = 1/2pKw + 1/2pKa + 1/2 log c

is completely wrong.

I've never seen that equation before, so I got out a pen and paper and tried to make sense of it, to me it looks OK,

1/2pKw + 1/2pKa + 1/2 log c = -1/2log[H⁺][OH^-] -1/2log[A^-][H⁺]/[AH] -1/2log[AH]
                                          = -1/2log[H⁺]² - 1/2log[H⁺]²
                                          = pH(water) + pH(the acid) = pH(total)

[Borek]

Assuming acetic acid 0.01M, Ka = 1.8*10^-5

Real pH about 3.39

Using this formula:

pH = 7 + 0.5*4.75 + 0.5*2 = 10.4

[Fragment_Error]

No, you need to watch the units! your concentration, 0.1M means 0.1 mol L^-1, so to find the number of moles, you MULTIPLY the concentration by the volume IN LITRES. Compare the units on each side, 
mol L^-1 = mol / L,
so mol = mol L^-1 x L

in other words, number of moles = concentration x volume (in litres)

Alright. I always get confused with that stuff. I can never remember if I should multiply or divide or whatever. So it's 0.1 x 0.0266 = 2.66 x 10^-3 Mol.

Write a balanced equation. (Hint: how many moles of NaCl will 1 mol of NaOH generate?)

I've got absolutley no idea.

K_a = [A^-][H⁺]/[AH]

example, for the equilibrium CH₃COOH <----> CH₃COO^-  + H⁺
AH = CH₃COOH
so A^- = CH₃COO^-

The square brackets mean 'concentration of'

You know [CH₃COOH]
You have worked out [CH₃COO^-] (it is the same as [CH₃COONa])
And if you look at the balanced equation I wrote above, you can see that the amount of CH₃COO^- produced is equal the amount of H⁺.

and therefore [CH₃COO^-]=[H⁺]

And therefore

K_a = [CH₃COONa]²/[CH₃COOH]

If you still can't get your head around this, I strongly suggest you discuss the concepts with your tutor. This is vey basic chemistry that you really should understand at degree level.

Ok, I know the concentration of CH₃COOH is 0.1M, the concentration of CH₃COONa is also 0.1M.

As you said, Ka = [CH3COONa]2/[CH3COOH]

Ka = 0.1²/0.1

This doesn't make any sense to me.


It's such basic chemistry, and I'm racking my brains just trying to figure out what the hell you guys are saying, let alone try to work out the problems. I didn't do chemistry at high school, and at university, it's assumed that all the students have done senior chemistry and passed, so they move very quickly through the work. I didn't even know what the hell a mole was until a few weeks ago, and this is like a foreign language to me.

Now, the formula you are given

pH = 1/2pKw + 1/2pKa + 1/2 log c

is completely wrong. For a weak acid (if some additional conditions are fulfilled) following formula works:

[H⁺ ] = sqrt(K_a*c)

It's for a strong acid and strong base: CH₃COOH vs NaOH

[Dan]

I've got absolutley no idea.

Look at your balanced equation!! 1 molecule of NaOH reacts with 1 molecule of HCl, how many molecules of NaCl do you get?

So how many molecules (in mol) do you get when 1 mol of NaOH reacts with 1 mol of HCl?

[Fragment_Error]

I've got absolutley no idea.

Look at your balanced equation!! 1 molecule of NaOH reacts with 1 molecule of HCl, how many molecules of NaCl do you get?

So how many molecules (in mol) do you get when 1 mol of NaOH reacts with 1 mol of HCl?

Oh man, I'm an idiot, sorry.

It's 2:19am here and am thinking very slowly.

HCl + NaOH -> NaCl + H₂O

1 molecule of HCl reacts with 1 molecule of NaOH to produce 1 molecule of NaCl.

Alright, I'm going to sleep now, and I'll check this thread out in the morning.

[Borek]

OK, I am back, I was in hurry writing previous answer so I have just posted number, without trying to analyze your post.

pH(water) + pH(the acid) = pH(total)

pH is not additive. You may add concentrations of H⁺ from different sources to find out total concentration (which is correct only if you assume that H⁺ from water autodissociation can be neglected), but for the above equation to be true you will need

log(a+b) = log(a) + log(b)

which is obviously wrong.

[Dan]

Oh christ, of course, sorry I was having real 'moment' there wasn't I?

I haven't really had to think about pH for a few years now, but even so, that was a massive schoolboy error!

[Borek]

Nobodys perfect

[Fragment_Error]

Ok, so part 2 I think I've fixed up now:

0.0266 x 0.1 = 2.66 x 10^-3


Now for concentration of CH₃COONa at the equivalence point:

pK_w = -log 10^-14
Ka = 4.75 (Gotten from this link)
c = 0.1??

For c, the concentration has to be in mol/L. How do I find that out?