# pKa and pKb calculation

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• February 5th 2008, 05:55 AM
Dan167
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 :D
• February 5th 2008, 11:34 AM
Jhevon
Quote:

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 :D

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
• February 5th 2008, 11:43 AM
earboth
Quote:

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
• February 5th 2008, 12:26 PM
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.
• February 5th 2008, 12:31 PM
Jhevon
Quote:

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
• February 5th 2008, 12:39 PM
bobak
Quote:

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
• February 5th 2008, 01:56 PM
Dan167
Wow i'm confused :o ... Thats all that is given I guess we have to make assumptions.... Does anyone mind going through it step by step :D
• February 5th 2008, 02:02 PM
Jhevon
Quote:

Originally Posted by Dan167
Wow i'm confused :o ... Thats all that is given I guess we have to make assumptions.... Does anyone mind going through it step by step :D

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
• February 5th 2008, 02:07 PM
Dan167
Quote:

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?
• February 5th 2008, 02:10 PM
Jhevon
Quote:

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 $x$
• February 5th 2008, 02:16 PM
Dan167
• February 5th 2008, 02:25 PM
Jhevon
Quote:

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

$x = \frac {-b \pm \sqrt{b^2 - 4ac}}{2a}$
• February 5th 2008, 02:27 PM
Dan167
Quote:

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.
• February 5th 2008, 02:38 PM
Dan167
Ok nope i'm lost .... :rolleyes:
• February 5th 2008, 02:40 PM
Jhevon
Quote:

Originally Posted by Dan167
Ok nope i'm lost .... :rolleyes:

ok, how about you post your attempt, so we can see where you're getting stuck and direct you
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