# pKa and pKb calculation

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• Feb 5th 2008, 02:43 PM
Dan167
rightyo I tried to put in values for the equation but I have no idea what goes where ... Don't worry I'll just make up the marks on the organic chemistry bit lol I give up on this stuff
• Feb 5th 2008, 02:54 PM
Jhevon
Quote:

Originally Posted by Dan167
rightyo I tried to put in values for the equation but I have no idea what goes where ... Don't worry I'll just make up the marks on the organic chemistry bit lol I give up on this stuff

i don't see your hang up, Dan.

you have $x^2 + K_ax - 0.1K_a = 0$ there are not many choices of where to put anything. you know that $K_a = 3.3 \times 10^{-2}$, so just replace $K_a$ with that wherever you see it, so you get

$x^2 + {\color{red}(3.3 \times 10^{-2})}x - 0.1{\color{red}(3.3 \times 10^{-2})} = 0$

so you get $x^2 + 0.033x - 0.0033 = 0$

so, $x = \frac {-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac {-0.033 \pm \sqrt{(0.033)^2 - 4(1)(0.0033)}}{2}$

and just plug that in your calculator

when you get the answer, take the logarithm of it and then take it's negative and you have the pH
• Feb 5th 2008, 02:58 PM
bobak
If your having difficulty with the qudratic formula just assume that the change in conceration of the acid can be nelgated and use the following formula instead.

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

your answer will be a bit off, but your chemistry teachers may not mind, you should ask them is that is allowed on your course (because it is on mine).

Yeah we have the same general feeling in my chemistry class as well most people make up all the marks on the organic chemistry, I however make up the marks with analytical chemistry ;) .
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