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<p class="MsoNormal"><tt><span style="font-size:10.0pt">I got some questions about buffers and what happens when strong acid or<o:p></o:p></span></tt></p>
<p class="MsoNormal"><tt><span style="font-size:10.0pt">base is added <o:p></o:p></span></tt></p>
<p class="MsoNormal"><tt><span style="font-size:10.0pt"><o:p> </o:p></span></tt></p>
<p class="MsoNormal"><tt><span style="font-size:10.0pt">A buffer is a special type of common-ion problem. It consists of the</span></tt><span style="font-size:10.0pt;font-family:"Courier New""><br>
<tt>conjugate acid-base pair of a weak acid and its conj. base (or weak base</tt><br>
<tt>and its conj. acid). Examples are:</tt><br>
<br>
<tt> HF/F-, CH3CO2H/CH3CO2-, NH3/NH4+</tt><br>
<br>
<tt>You could get a buffer by simply adding both members of the weak conj.</tt><br>
<tt>acid-base pair. You could also get it by starting with a weak acid</tt><br>
<tt>and adding some strong base to convert some of the weak acid to its</tt><br>
<tt>conj. base but leaving some of the weak acid behind (such as starting</tt><br>
<tt>with HF and adding some NaOH to react with the HF to produce F- but</tt><br>
<tt>leave some HF). You could start with NH3 and add some strong acid to</tt><br>
<tt>convert some of the weak NH3 to its conj. acid, NH4+, leaving some of</tt><br>
<tt>the NH3 and so you wind up with both NH3 and NH4+ in solution (like<o:p></o:p></tt></span></p>
<p class="MsoNormal"><tt><span style="font-size:10.0pt">the example I did in lecture).</span></tt><span style="font-size:10.0pt;font-family:"Courier New""><br>
<br>
<tt>A strong acid with a strong base (HCl/NaOH) is NOT a buffer system.<o:p></o:p></tt></span></p>
<p class="MsoNormal"><tt><span style="font-size:10.0pt">A strong acid with it’s conjugate base is not a buffer. A strong base<o:p></o:p></span></tt></p>
<p class="MsoNormal"><tt><span style="font-size:10.0pt">with it’s conjugate acid won’t be a buffer (like NaOH and Na+).</span></tt><span style="font-size:10.0pt;font-family:"Courier New""><br>
<br>
<tt>For buffers you want to remember, the Henderson-Hasselbalch eqn. It</tt><br>
<tt>can be used in two ways. It's technically,</tt><br>
<br>
<tt> [base]<sub>eq</sub> </tt><br>
<tt>pH = pKa + log -----------</tt><br>
<tt> [acid]<sub>eq</sub> </tt><br>
<br>
<tt>where the ratio gives the equilibrium conc. of the two components of</tt><br>
<tt>the buffer (the weak conj. acid-base pair).</tt><br>
<br>
<tt>You can use the above eqn if you're given the pH and pKa and asked for<o:p></o:p></tt></span></p>
<p class="MsoNormal"><tt><span style="font-size:10.0pt">the ratio of the conc. of the base and acid in the solution or you're<o:p></o:p></span></tt></p>
<p class="MsoNormal"><tt><span style="font-size:10.0pt">trying to make a buffer at a specific pH and you need to figure out<o:p></o:p></span></tt></p>
<p class="MsoNormal"><tt><span style="font-size:10.0pt">how much of one of the components has to be added.<o:p></o:p></span></tt></p>
<p class="MsoNormal"><span style="font-size:10.0pt;font-family:"Courier New""><br>
<br>
<tt>The above eqn can also be written as:</tt><br>
<br>
<tt> [base]<sub>orig</sub> + x </tt><br>
<tt>pH = pKa + log ----------------</tt><br>
<tt> [acid]<sub>orig</sub> - x </tt><br>
<br>
<tt>where [base]<sub>eq</sub> = [base]<sub>orig</sub> + x and [acid]<sub>eq</sub> = [acid]<sub>orig</sub> – x . This comes<o:p></o:p></tt></span></p>
<p class="MsoNormal"><tt><span style="font-size:10.0pt">from the ICE table you could use instead of this eqn.</span></tt><span style="font-size:10.0pt;font-family:"Courier New""><br>
<br>
<tt>If the original conc. of acid and base are relatively large, their ratio</tt><br>
<tt>is between 0.1 and 10 and the Ka is relatively small we ignore the "x"</tt><br>
<tt>in both the numerator and denominator to get the eqn below using just<o:p></o:p></tt></span></p>
<p class="MsoNormal"><tt><span style="font-size:10.0pt">the original conc. given in the problem.</span></tt><span style="font-size:10.0pt;font-family:"Courier New""><br>
<br>
<tt> [base]<sub>orig</sub> </tt><br>
<tt>pH = pKa + log -----------</tt><br>
<tt> [acid]<sub>orig</sub> </tt><br>
<br>
<br>
<tt>If the above conditions aren't true you wouldn't be able to ignore the "x"</tt><br>
<tt>and you would have to set up an ICE table and solve a quadratic. See<o:p></o:p></tt></span></p>
<p class="MsoNormal"><tt><span style="font-size:10.0pt">Sample Exercise 17.4 in the textbook for an example of this.</span></tt><span style="font-size:10.0pt;font-family:"Courier New""><br>
<br>
<tt>Finding the pH of a buffer soln given conc. for the base and acid is just</tt><br>
<tt>a matter of plugging in numbers in this eqn.</tt><br>
<br>
<tt>The problem comes in when adding a strong acid or strong base to the buffer</tt><br>
<tt>system. The discussion below is for a buffer of a weak acid HA and its</tt><br>
<tt>conj. base, A-. The equilibrium for the ICE table would be</tt><br>
<br>
<tt> HA <==> H+ + A-</tt><br>
<br>
<tt>Again, you could likely use the HH eqn and not have to set up the ICE table.</tt><br>
<br>
<tt>You need to understand what happens when you add strong acid or base to the</tt><br>
<tt>system. This seems to be the main question I'm getting. Here's what happens</tt><br>
<tt>in a nutshell.</tt><br>
<br>
<tt>Strong Acid - reacts with the A- (base) part of the buffer system to create</tt><br>
<tt> additional HA and leaves some A-. You'll still have a buffer</tt><br>
<tt> as long as the amount of strong acid added was small (in terms</tt><br>
<tt> of the moles added). You need to do a stoichiometry table for</tt><br>
<tt> the neutralization reaction between the strong acid and A-.</tt><br>
<tt> This is a BCA table (the ICC tables I used in lecture), in MOLES.</tt><br>
<tt> Using H+ to stand for the strong acid, this is the net ionic<o:p></o:p></tt></span></p>
<p class="MsoNormal" style="text-indent:.5in"><tt><span style="font-size:10.0pt"> eqn needed,</span></tt><span style="font-size:10.0pt;font-family:"Courier New""><br>
<br>
<tt> H+ + A- ==> HA (in moles)</tt><br>
<br>
</span><br>
<tt><span style="font-size:10.0pt">Strong Base - reacts with the HA (acid) part of the buffer system to create</span></tt><span style="font-size:10.0pt;font-family:"Courier New""><br>
<tt> additional A- and leaves some HA. You'll still have a buffer</tt><br>
<tt> as long as the amount of strong base added was small (in terms</tt><br>
<tt> of the moles added). You need to do a stoichiometry table for</tt><br>
<tt> the neutralization reaction between the strong base and HA.</tt><br>
<tt> This is a BCA table (the ICC tables I used in lecture), in MOLES.</tt><br>
<tt> Using OH- to stand for the strong base this is the net ionic<o:p></o:p></tt></span></p>
<p class="MsoNormal" style="text-indent:.5in"><tt><span style="font-size:10.0pt"> eqn needed,</span></tt><span style="font-size:10.0pt;font-family:"Courier New""><br>
<br>
<tt> HA + OH- <==> A- + H2O (in moles)</tt><br>
<br>
<tt>Once you do the proper neutralization reaction you can plug the new conc.</tt><br>
<tt>of A- and HA back into the HH eqn and calculate the new pH. Technically,</tt><br>
<tt>if you remember from class, you can use the moles of A- and HA you get from<o:p></o:p></tt></span></p>
<p class="MsoNormal"><tt><span style="font-size:10.0pt">doing the above neutralization reactions since they're in the same volume of<o:p></o:p></span></tt></p>
<p class="MsoNormal"><tt><span style="font-size:10.0pt">soln and the ratio of moles is the same as the ratio of conc.,</span></tt><span style="font-size:10.0pt;font-family:"Courier New""><br>
<br>
<tt> [A-] (mol A-/V_total) mol A-</tt><br>
<tt> ------ = ------------------ = --------</tt><br>
<tt> [HA] (mol HA/V_total) mol HA</tt><br>
<br>
<br>
<tt><o:p></o:p></tt></span></p>
<p class="MsoNormal"><tt><span style="font-size:10.0pt">That's about it. I hope this helps.</span></tt><span style="font-size:10.0pt;font-family:"Courier New""><br>
<br>
<tt>Dr. Zellmer</tt></span><span style="font-family:"Arial",sans-serif"><o:p></o:p></span></p>
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