# What is the ph of .035M acetic acid ?

1. Sep 19, 2004

### vikki99

can someone give me some insight to solve this
in detail :

what is the ph of .01 Naoh ?
and what is the ph of .035M acetic acid ?

2. Sep 20, 2004

### chem_tr

Hello,
The first question can easily be solved, since NaOH fully dissociates. But I can not say the same for acetic acid, you'll need to give the dissociation constant for it. I remember that it is $$1,8*10^{-5} mol^2L^{-2}$$, and will solve by using this value. If I'm mistaken, please replace it with the correct one.

Firstly, let's look at NaOH:

$$\underbrace{NaOH} \rightarrow \underbrace{Na^+} + \underbrace{OH^-}$$
$$\overbrace{0,01-x} \rightarrow \overbrace{x} + \overbrace{x}$$

Since dissociation constant for NaOH is extremely big, we never need to calculate the $${0,01-x}$$; it has a very very close value to 0,01. So we can easily assume that the $$[OH^-]$$ is 0,01, the same goes for $$[Na^+]$$.

As pH is the negative logarithm of $$[H^+]$$ or $$\frac{10^{-14}}{[OH^-]}$$, we'll use whichever is suitable. Since we know the hydroxide concentration, let's use this:

$$pH=-\log(\frac{10^{-14}}{0,01})=12$$.

For acetic acid, we'll consider the $$x$$ values as it is not dissociated 100% in water.

$$\underbrace{CH_3COOH} \rightarrow {CH_3COO^-}+{H^+}$$
$$\overbrace{(0,035-x)}\rightarrow$$

We are given that $$\frac{x^2}{(0,035-x)}=1,8*10^{-5}$$. Then it's easy to find x, either by omitting it or by solving a two-unknown equation with $$\Delta=b^2-4ac$$ and $$x_1=\frac{-b-\sqrt{\Delta}}{2a}$$ and $$x_2=\frac{-b+\sqrt{\Delta}}{2a}$$. Note that only one root gives a valid value, just omit the other.

I will not consider x, if you really wonder, you may not omit it and solve the two-unknown equation. When we omit it, we'll find that $$x=\sqrt{0,035*1,8*10^{-5}}=7,94*10^{-4}$$, hence we find the pH to be $$3,1$$.

Regards,
chem_tr

3. Sep 21, 2004

### vikki99

thank chem_tr I have one more question :

Solve the quadratic equation for the more general case where the total concentration of acid is:

a. ((Ao = (HA) + (A-) ) has any value Ao.
b. One can avoid solution of the quadratic equation by using a method of successive approximations, starting with x2 =AoKa.
Explain how this would be done.

thanks again for the help earlier

4. Sep 21, 2004

### chem_tr

Hello,

I will not be as "helpful" as I did before, since it may not help you as I intended; I don't want to be harmful for your education. I decided to show you the way instead, that's better for you I think.

The total is the initial concentration of the acid. So you disregard how much of it is ionized; just give the initial concentration, and it is over. It is your task to show it mathematically.

If H+ and A- are said to be ionized as much as x, then you may easily write the equilibrium constant using C0, Ka, and x2, along with my earlier posts of course

Best wishes and have a good study,
chem_tr

5. Sep 21, 2004

### vikki99

THANK YOU CHEM_tr

thanks soooooooooooooooooooooooo much for your help
you are not ruining my education
I dont have alot of support here at the school im attending at the only tutor is a bit "special " i mean slow.
thanks again