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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.