Moderators: Chem_Mod, Chem_Admin

alexis castro 1B
Posts: 30
Joined: Wed Nov 16, 2016 3:02 am


Postby alexis castro 1B » Sun Jul 30, 2017 12:59 pm

A buffer solution of volume 100.0 mL is 0.100 m CH3COOH(aq) and 0.100 m NaCH3CO2(aq). (a) What are the pH and the pH change resulting from the addition of 10.0 mL of 0.950 m NaOH(aq) to the buffer solution? (b) What are the pH and the pH change resulting from the addition of 20.0 mL of 0.100 m HNO3(aq) to the initial buffer solution?

Can someone show me how to setup up this problem? I'm confused on how to apply the henderson-hasselbalch equation.

Ben Rolnik 1D
Posts: 33
Joined: Fri Jun 23, 2017 11:39 am

Re: 13.11

Postby Ben Rolnik 1D » Sun Jul 30, 2017 11:41 pm

This question is really tricky (unless you actually go through the Toolboxes and examples given in chapter 13 in the book).

but for (a) basically you know that for the equilibria equation, "HA - X" and "A + X" (from your ice box) is basically going to be the same as HA and X.... therefore you can use the equation: pH = pKa + log (Molarity A / Molarity HA).

To finish the second part of part (a) and also to complete part (b) you need to understand

1) For pH change you need to take the Moles Analyte minus the Moles Titrant divided by VOLUME TOTAL (analyte plus titrant)... then you ALSO need to consider "moles titrant divided by VOLUME TOTAL" and add that to your icebox / equilibria equation.

2) AFTER the stoichiometric point, you invert the equation so instead of solving for the concentration of H+ you solve for OH-.... e.g. CO2 - + H20 --> CooH + OH - .....

With that information you should be able to solve this :)

Return to “*Making Buffers & Calculating Buffer pH (Henderson-Hasselbalch Equation)”

Who is online

Users browsing this forum: No registered users and 1 guest