# Math Help - pKa and pKb calculation

1. ## pKa and pKb calculation

Hey, anyone know how I would work this out?

Calculate the pH of 0.1M dichloroacetic acid, Ka = 3.3 x 10-2.

Thanks

2. Originally Posted by Dan167
Hey, anyone know how I would work this out?

Calculate the pH of 0.1M dichloroacetic acid, Ka = 3.3 x 10-2.

Thanks
we are given little to work with here, we have to go the long way around. (unless you are dealing with a buffer and you neglected to mention the molarity of the basic part of the acid). what is the formula for dichloroacetic acid? we have to write an equation for the ionization reaction using it

3. Originally Posted by Jhevon
we are given little to work with here, we have to go the long way around. (unless you are dealing with a buffer and you neglected to mention the molarity of the basic part of the acid). what is the formula for dichloroacetic acid? we have to write an equation for the ionization reaction using it
Hi,

have a look here: Dichloroacetic acid - Wikipedia, the free encyclopedia

4. well we have enough information to solve his (providing we make a few assumptions)

$Ka = \frac{[H+] \cdot [salt]}{[Acid]}$

if we assume that the solution is not buffered then $[H+] = [Salt] = x$

so then $Ka = \frac{x^2}{[Acid] - x }$

and there you have enough information to solve the equation.

5. Originally Posted by bobak
well we have enough information to solve his (providing we make a few assumptions)

$Ka = \frac{[H+] \cdot [salt]}{[Acid]}$

if we assume that the solution is not buffered then $[H+] = [Salt] = x$

so then $Ka = \frac{x^2}{[Acid] - x }$

and there you have enough information to solve the equation.
indeed. i wanted to write out the ionization reaction just to be sure. it been a while since i did this. we have that $pH = - \log x$ if that is the equation for the Ka. the [Acid] - x seems weird to me, but as i said, it's been a while

6. Originally Posted by Jhevon
the equation for the Ka. the [Acid] - x seems weird to me, but as i said, it's been a while
well in chemistry class we are usually allowed to assume that the change in concentration of the acid may be neglected as it is so small so most people just leave the denominator as [Acid]. however this assumption is only really valid for very small values in Ka, but chemistry teachers don't really want to waste time teaching students how to solve quadratic equations so they usually make the assumption to allow easier computation of the ph

7. Wow i'm confused ... Thats all that is given I guess we have to make assumptions.... Does anyone mind going through it step by step

8. Originally Posted by Dan167
Wow i'm confused ... Thats all that is given I guess we have to make assumptions.... Does anyone mind going through it step by step
we are told [Acid] = 0.1 M. use this to solve for x in the equation (which you should be familiar with) that bobak posted. once you have x, pH = -log(x). that's it

9. Originally Posted by Jhevon
we are told [Acid] = 0.1 M. use this to solve for x in the equation (which you should be familiar with) that bobak posted. once you have x, pH = -log(x). that's it
Lol i'm not familiar with it ... how do you solve it for x?

10. Originally Posted by Dan167
Lol i'm not familiar with it ... how do you solve it for x?
you have $K_a = \frac {x^2}{0.1 - x}$

$\Rightarrow K_a(0.1 - x) = x^2$

$\Rightarrow x^2 - K_a(0.1 - x) = 0$

$\Rightarrow x^2 + K_ax - 0.1K_a = 0$

plug in the value for $K_a$, you will get a quadratic equation. then just use the quadratic formula to solve for $x$

12. Originally Posted by Dan167
given a quadratic equation (in the variable x) of the form $ax^2 + bx + c = 0$

we can find the solutions to this equation by using the formula

$x = \frac {-b \pm \sqrt{b^2 - 4ac}}{2a}$

13. Originally Posted by Jhevon
given a quadratic equation (in the variable x) of the form $ax^2 + bx + c = 0$

we can find the solutions to this equation by using the formula

$x = \frac {-b \pm \sqrt{b^2 - 4ac}}{2a}$
Right .... I think i've got it, thanks.

14. Ok nope i'm lost ....

15. Originally Posted by Dan167
Ok nope i'm lost ....
ok, how about you post your attempt, so we can see where you're getting stuck and direct you

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