Calculating pH of buffer after adding HCl

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SUMMARY

The discussion focuses on calculating the pH of a Tris buffer after the addition of 1 ml of 0.05M HCl. Initially, the buffer has a pH of 9.0, with 4 ml of 0.01M Tris. Using the Henderson-Hasselbalch equation, the ratio of base to acid was determined to be 6.17. After adding HCl, the calculations indicate that all base has been converted to the acid form, resulting in excess HCl that must be accounted for to find the new pH, as the Henderson-Hasselbalch equation becomes inapplicable.

PREREQUISITES
  • Understanding of the Henderson-Hasselbalch equation
  • Knowledge of buffer solutions and their components
  • Basic chemistry concepts related to pH and molarity
  • Familiarity with Tris buffer and its properties
NEXT STEPS
  • Study the calculation of pH changes in buffer solutions upon acid or base addition
  • Learn about the properties and applications of Tris buffer in biochemical experiments
  • Explore alternative methods for calculating pH when the Henderson-Hasselbalch equation is not applicable
  • Research the concept of buffer capacity and its significance in solution chemistry
USEFUL FOR

Chemistry students, laboratory technicians, and anyone involved in buffer solution preparation and pH calculations will benefit from this discussion.

MissSpice
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Homework Statement


If the Tris buffer was exactly pH=9.0, calculate expected pH value after addition of 1 ml of 0.05 HCl.

Buffer: 4 ml of 0.01M Tris, pH 9.0
HCl: 1 ml of 0.05M HCl

Homework Equations


H-H: pH=pKa+log[A-]/[HA]

The Attempt at a Solution


9.0=8.21+log[A-]/[HA]
[A-]/[HA]=6.17

[A-]+[HA]=0.01M so, [A-]=6.17[HA], and 6.17[HA]+[HA]=0.01
7.17[HA]=0.01, so [HA]=.0013M
0.01-.0013=.0087M=[A-]

HCl: 1ml x 0.05M= 0.05mmols HCl

[A-]=.0087M x 4 ml= .0348mmol A-
[HA]=.0013M x 4 ml= .0052 mmol HA

Here's the part I'm unsure of:
If I plug these values into the Henderson-Hasselbalch, I'm going to get a negative value for the base part of the equation. I'm aware that this means that all of the base has been converted to the acid form, and that there is actually HCl left over that contributes to the pH. I'm just not sure how to calculate the pH from here, since the H-H equation can't be used.
 
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Assuming your calculations were right - just from the excess HCl.
 

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