Talk:RICE chart

Needs some qualifiers
There's no mention of the restricted applicability of this method for calculating pH. There's an underlying assumption that [H+] >> [OH-]. Another way of saying this is that this only applies to solutions whose pH is much less than 7. Klortho (talk) 05:34, 13 March 2011 (UTC)

SOMEONE DO SOMETHING ABOUT THIS PAGE
Someone who knows HTML needs to create an ICE table to demonstrate to others what exactly is an ICE table... --XH 20:55, 20 January 2007 (UTC)User:Xinyu

I've now done this, basically by paraphrasing the material below. I hope LestatdeLioncourt does not mind. Petergans 19:05, 17 May 2007 (UTC)

pH value from dissociation constant
The following was from the March 12 Science Reference Desk.


 * This is based on a homework question with specific values, but I am asking only for general formula.

If I know the acid dissociation constant (pKa) and the concentration of a weak acid, how would I find the pH? How do I find the pH after adding a certain volume of a strong base? − Twas Now ( talk • contribs • e-mail ) 00:58, 12 March 2007 (UTC)

Take a look at Henderson-Hasselbalch equation
 * $$\textrm{pH} = \textrm{pK}_{a}+ \log_{10} \frac{[\textrm{A}^-]}{[\textrm{HA}]}$$

From the pKa page the ionised acid and H+ concentration will be the same, so you end up with pH+pH=-log10(concentration of a weak acid)+pKa.
 * $$\textrm{pH} = (\textrm{pK}_{a}- \log_{10} {[\textrm{HA}]}) / 2$$

GB 02:18, 12 March 2007 (UTC)


 * Thanks. I guess what confuses me is how do I determine [HA] and [A-] for a given weak acid (and its conjugate base)? − Twas Now ( talk • contribs • e-mail ) 03:09, 12 March 2007 (UTC)

It's really very simple. All you need to do is set up an ICE table for the ionization equilibrium of the acid, i.e. HA &#8652; A- + H+ :

Because $$pH = -\log{[H^+]}$$, and you can tell from the table that when the ionization is over [H+] = x, then $$pH = -\log{x}$$. So, your job is to find x.

The equilibrium constant expression for the ionization is:

$$K_a = \frac{[H^+][A^-]}{[HA]}$$

Substitute the concentrations with the values found in the last row of the ICE table.

$$K_a = \frac{x.x}{c_0 - x}$$

Now plug in the specfic values for c0 and Ka ($$K_a = 10^{-pK_a}$$) provided in your question, then solve for x, and you're done!

If you do some math, you can quickly figure out that the relation GB provided is only a different (and handier) way of expressing Ka, especially when you're working with buffer solutions.

Now for the second part of your question: when you added a certain quantity of a strong base, the OH- ions you just added will react with some of the H+ in solution. From the OH- - H+ reaction stoichiometry, you can calculate the number of moles of H+ that have disappeared and how many moles are left. Then you can find the new concentration of H+ ions (keep in mind the change in volume) and the new pH. &mdash;  Lestat deLioncour t  14:45, 12 March 2007 (UTC)

Typesetting
Terrible (and inconsistent) typesetting in this article. Needs to be fixed. —DIV (1.129.104.52 (talk) 12:02, 18 August 2019 (UTC))