lecture+4.105

lecture+4.105 - Biological Sciences 105 Lecture 4, January...

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Biological Sciences 105 Lecture 4, January 13, 2011 Copyright Steven M. Theg, 2011. All federal and state copyrights reserved for all original material presented in this course through any medium, including lecture or print. 1 LAST TIME WE ENDED WITH THE AN EXAMPLE PROBLEM QUIZ, THEN SOME QUICK PROBLEMS DON’T DO THESE IN CLASS UNLESS THERE ARE SPECIFIC QUESTIONS. Example #1 : In a 1.0 L solution of 0.1 M Tris at pH 7.5, what is the concentration of protonated and unprotonated Tris? The solution comes from these steps: 1. Determine [A - ]/[HA] from the H-H equation 2. Express [A - ] as x [HA] 3. C T = [A - ] + [HA] = x [HA] + [HA] = (1+x)[HA] 4. Plug in numbers 5. Check that it makes sense Use Henderson-Hasselbalch equation pH = pK a + log{[A - ]/[HA]} 7.5 = 8.1 + log{[A - ]/[HA]} log{[A - ]/[HA]} = 7.5 - 8.1 = -0.6 [A - ]/[HA] = 10^(-0.6) = 10 -0.6 = 0.25 [A - ] = 0.25[HA] C T = total concentration = [A - ] + [HA] 0.1 = 0.25[HA] + [HA] = 1.25[HA] [HA] = 0.1/1.25 = 0.08 M [A - ] = 0.1 - 0.08 = 0.02 M This answer makes sense since pH < pK a , and so should have more protonated than unprotonated buffer. Example #2 : How many moles of Tris base and HCl are required to make a 1.0 L solution of 0.1 M Tris-HCl at pH 7.8?
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Biological Sciences 105 Lecture 4, January 13, 2011 Copyright Steven M. Theg, 2011. All federal and state copyrights reserved for all original material presented in this course through any medium, including lecture or print. 2 Solution from these steps: 1. Solve for A - /HA using H-H equation. 2. Solve for A - = xHA 3. Replace A - with xHA in C T = A - + HA 4. Solve for numerical values of HA and A - 5. Need as many moles of acid as conjugate acid produced from conjugate base. 6. Check that the answer makes sense. For this example, pH = pKa + log(A - /HA) 7.8 = 8.1 + log(A - /HA) A - /HA = 10 -0.3 = 0.5 A - = 0.5 × HA C T = 0.1 = A - + HA = 0.5 × HA + HA = 1.5 × HA HA + 0.1/1.5 = 0.067 So we need 0.067 moles of HCl to convert 0.067 of A - into HA. Example #3 : How many moles of Tris base and HCl are required to make a 1.0 L solution of 0.1 M Tris-HCl at pH 8.1? This can be solved by inspection. Since the pH is the pK a , we need to neutralize half of the Tris. So we need 0.1 moles of Tris base and 0.05 moles of HCl. Example #4: Add 0.005 moles H + per liter to 20 mM Tris at pH 7.8, what is the final pH? Solution from these steps: 1. What are the initial concentrations of HA and A - ? Get these from the Henderson- Hasslebalch equation. 2. Calculate [HA] i + 0.005 mol/L H + . That’s [HA] f . 3. From that calculate [A - ] f . 4. Put the new ratios into the Henderson-Hasslebalch equation and calculate pH f . 5. Does this make sense? 6. Answer I got is pH = 7.06 pH = pKa + log(A - /HA) 7.8 = 8.1 + log(A - /HA) A - /HA = 10 -0.3 = 0.5 A - = 0.5 × HA
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Biological Sciences 105 Lecture 4, January 13, 2011 Copyright Steven M. Theg, 2011. All federal and state copyrights reserved for all original material presented in this course through any medium, including lecture or print. 3
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This note was uploaded on 01/18/2011 for the course BIS 105 taught by Professor Kennethhilt during the Spring '09 term at UC Davis.

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lecture+4.105 - Biological Sciences 105 Lecture 4, January...

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