How to use equations to show how a buffer resists changes in pH.?

How do I show equations on how a lactate buffer resists changes in pH? The components of the lactate buffer are lactic acid, HC3H5O3, and lactate ion, C3H5O3-, often provided as sodium salt.



I know it's probably very simple, but I can't find an example, so I'm not sure exactly what it's asking. Thanks so much!How to use equations to show how a buffer resists changes in pH.?
The acid is actually in an aqueous solution which means initially, it is presented in water.



Then the acid and water reacts to create H3O+ and C3H5O3-. When you add in sodium lactate ion, NaC3H5O3-, you are basically adding in C3H5O3- to the product side into the C3H5O3- from the reaction of acid+water.



This means that the equilibrium will shift back to the left since you add in more products. When it shifts back, it sort of cancels out the reactions going right caused by the reactants hence, not losing any H+ ions, hence resists change in pH.



Now to write it out:



HC3H5O3(aq) + H2O(l) %26lt;==%26gt; H+(aq) + C3H5O3-(aq)



note: H+ = H3O+ and the Na salt is a spectator ion so we don't write it out.



And the Equilibrium equation would look like:



Kc = [C3H5O3-][H+]/[HC3H5O3]

= (C3H5O3- + x)(x)/(HC3H5O3-x)



Which can be simplified as:



Kc = (C3H5O3-)(x)/(HC3H5O3)

note: we assume x is very small so we take it out and the reason it is small is because in buffer we add x to to product side to cancel out the x in the reactant side.



Look, I know it's confusing, just take time to practice these problems and ask questions when you are stuck and you will succeed. Good Luck.How to use equations to show how a buffer resists changes in pH.?
You can easy to calculate pH of the buffer by using the

For the buffer solution you can use the Henderson-Hasselbach equation:



pH = pKa + log ([A-] / [HA] )



according that equation above pH of the buffer is determined by the ratio of ([A-] / [HA] ) so we must concern with this ratio to show how a buffer resists changes in pH.



The concentration of A- can changes if we add acid to the buffer solution as this reaction:



C3H5O3Na + H+ -%26gt; HC3H5O3 + Na+



This action makes the ratio of ([A-] / [HA] decrease, because the concentration of HC3H5O3 increases.



And the concentration of HA can changes too if we add base to the solution as this reaction



HC3H5O3 + OH- -%26gt; C3H5O3- + H2O





This action makes the ratio of [A-] / [HA] increase because the concentration of C3H5O3- increases.



As thsese examples



[C3H5O3-] / [HC3H5O3] initial = (0.50)/(0.50) = 1

When you add acid to the buffer = (0.49)/(0.51)= 0.96

When you add base to the buffer = (0.51)/(0.49)=1.04





The change in the ratio [C3H5O3-] / [HC3H5O3] is very small thus the [H+] and the pH remains constant.



hopely my answer can help you...
All you can really put is:



Ka = ([C3H5O3-][H+] )/ [HC3H5O3]



pH = -log [H+]



[H+] = [C3H5O3-] / (Ka x [HC3H5O3])



pH = - log ([C3H5O3-] / (Ka x [HC3H5O3]))



I think that should do.
HC3H5O3 ---%26gt; C3H5O3- + H+

If you look up the Ka value of lactic acid you will notice it is small. This means that the vast majority of its molecules in solutiona re undissociated. Hardly any molecules ionise to form lactate ions and hydrogen ions. So [HC3H5O3] is high.



The Sodium lactate is a soluble salt. it ionises completely.

C3H5O3-Na+---%26gt; C3H5O3- + Na+

So [C3H5O3-] is high.



If we add acid to a buffer solution, the solution counteracts this by removing them.

HC3H5O3 === C3H5O3- + H+

The position of the equilibrium shifts to the left. Sodium lactate provides the solution with lactate ions which 'mop up' the H+ ions added. Forming lactic acid and resisting changing in pH.



If we add alkali, OH- ions, these react with any H+ ions in the solution (which come from the lactic acid dissociating).

H+ + OH- --%26gt; H20

This shift the equilibrium to the right to ''replace' H+ ions.

We have plent of undisociated HC3H5O3 molecules available because its a weak acid. These are ready to split up and replace the lost H+ ions.



(In this --%26gt; are normal arrows, and === are equilibrium arrows)

Hope this helps.
This is a typical exam question, and all they require is you to show lactic acid reacting with OH-, and lactate reating with H+.



Two balanced equations...