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 [tex]1,8*10^{-5} mol^2L^{-2}[/tex], and will solve by using this value. If I'm mistaken, please replace it with the correct one.

Firstly, let's look at NaOH:

[tex]\underbrace{NaOH} \rightarrow \underbrace{Na^+} + \underbrace{OH^-}[/tex]
[tex]\overbrace{0,01-x} \rightarrow \overbrace{x} + \overbrace{x}[/tex]

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

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

[tex]pH=-\log(\frac{10^{-14}}{0,01})=12[/tex].

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

[tex]\underbrace{CH_3COOH} \rightarrow {CH_3COO^-}+{H^+}[/tex]
[tex]\overbrace{(0,035-x)}\rightarrow[/tex]

We are given that [tex]\frac{x^2}{(0,035-x)}=1,8*10^{-5}[/tex]. Then it's easy to find x, either by omitting it or by solving a two-unknown equation with [tex]\Delta=b^2-4ac[/tex] and [tex]x_1=\frac{-b-\sqrt{\Delta}}{2a}[/tex] and [tex]x_2=\frac{-b+\sqrt{\Delta}}{2a}[/tex]. 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 [tex]x=\sqrt{0,035*1,8*10^{-5}}=7,94*10^{-4}[/tex], hence we find the pH to be [tex]3,1[/tex].

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