Please help me find the pH of a solution.

[ddatuf]

I am having trouble doing these two problems. They are about finding the pH in solutions. Here they are:

 

a) 1 liter of 0.1M acetic acid (pKa=4.7) to which has been added 75 ml of 1M NaOH.

 

b) 1.0 liter of 0.1M glycine hydrochloride (pKa values are 2.3 and 9.7) to which has been added 100 ml of 1M NaOH.

 

The pH must be determined for these problems. Any help would be much appreciated. Thank you.

[John Cuthber]

http://en.wikipedia.org/wiki/Henderson_hasselbach

[Riogho]

Use logs and antilog functions

 

And if that fails, there is always titration

[thedarkshade]

Riogho is right!

 

You need to know this too:

 

[math]Kw=[OH^-][H^+]=[1\times 10^{-7}\frac{mol}{dm^3}][1\times 10^{-7}\frac{mol}{dm^3}][/math]

 

[math]Kw=1\times 10^{-14}\frac{mol^2}{dm^6}[/math]

 

Not try to find pH using antilog that Riogho mentioned!

 

Note: pH represent the concentration of [math]H^+[/math] ions!

 

edit: swansont don't remove my post please, it's not the answer:D!

 

Cheers,

Shade

[John Cuthber]

Tehnically, you need to consider the effect if the self ionisation of water which is where the Kw comes from but it will be a small effect, you can probably ignore it unless you need a very high accuracy (rather more accuracy than the Pka values you have been given).

You do need to calculate logs (rather than antilogs) but I can't see what use a titration would be. If Riogho would care to enlighten me on that I'd be happy to here about it; otherwise perhaps he might care not to complicate the issue.

 

Ddatuf,

did you understand the stuff on that wiki page?

It tells you how to do the calculation, but, like a lot of things, it might not be a very clear explanantion.

[thedarkshade]

I think he means negative log when saying antilog!

[John Cuthber]

Possibly, but if he doesn't know what he's talking about...

[ddatuf]

I actually tried the problem out recently and got an answer. My answer was different from the correct one by 2 tenths of a decimal place. Is that an incorrect? I know that a difference in pH of 5 and 6 is big, but what about the little differences like that?

[John Cuthber]

It's difficult to know without seeing what you did.

I think that not taking account of the volume would make a difference about that big.

If you add 75ml of NaOH soln to 1 litre of Acid you end up with 1.075 litres; if you forgot about that and just assumed a final volume of 1 litre I think the difference would be about that big.

On the other hand, I guess it's easier to ask whoever set the question.

[ChemSiddiqui]

I happened to look at his pos and just want to clear something. Antilog of x is actually 10 ^x .

[ddatuf]

Hey thanks guys. I went to a teacher and he helped me get the answer. I really appreciate the help though.

[thedarkshade]

I happened to look at his pos and just want to clear something. Antilog of x is actually 10 ^x .

It's not really literally anti, like the inverse function. It's just the negative logarithm, as follows:

 

[math]pH=-log10^{-7}[/math]

[math]pH=-(-7)log10[/math]

[math]pH=-(-7)[/math]

[math]pH=7[/math]

 

You get the same for pOH too. That's why pH + pOH = 14 (or [math]-log10^{-14}[/math])!