What is the ph of .035M acetic acid ?

[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 ?

 

[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

 

[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

 

[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, K_a, and x², along with my earlier posts of course

Best wishes and have a good study,
chem_tr

 

[vikki99]

THANK YOU CHEM_tr

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