help please

[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]

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, Ka, and x2, along with my earlier posts of course :smile:

Best wishes and have a good study,
chem_tr

[vikki99]

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