Asked by Sara
One liter of a 0.1 M Tris buffer (pKa of Tris = 8.3, ) is prepared and adjusted to a pH of 2.0.
[HA] = 0.10 and [A-] = 5.012E-8
What is the pH when 1.5 mL of 3.0 M HCl is added to 1.0 L of the buffer?
[HA] = 0.10 and [A-] = 5.012E-8
What is the pH when 1.5 mL of 3.0 M HCl is added to 1.0 L of the buffer?
Answers
Answered by
DrBob222
Sara, do you have answer choices? The answer I'm getting doesn't seem realistic so I may be missing something.
Answered by
Sara
I have no answer choices. My friend had the same problem, but different pH. Her problem is:
One liter of 0.10 M Tris buffer (pKa = 8.30) is prepared and adjusted to pH 8.2.
[A⁻]=0.044M and [HA]=0.056M
What is the pH when 1.5 mL of 3.0 M HCl is added to 1.0 L of the buffer?
The pH she got was 8.1.
I did the same work as her, but I keep getting the wrong answer.
One liter of 0.10 M Tris buffer (pKa = 8.30) is prepared and adjusted to pH 8.2.
[A⁻]=0.044M and [HA]=0.056M
What is the pH when 1.5 mL of 3.0 M HCl is added to 1.0 L of the buffer?
The pH she got was 8.1.
I did the same work as her, but I keep getting the wrong answer.
Answered by
DrBob222
For your friends problem I get 8.11.
mmols A^- = 0.044 x 1000 = 44
mmols HA = 0.056 x 1000 = 56
mmols HCl added = 1.5 mL x 3.0 M = 4.5, then
..............A^- + H^+ ==> HA
I............44.......0............56
add..................4.5......................
C..........-4.5....-4.5........+4.5
E..........39.5.......0...........60.5
pH = 8.3 + log (39.5/60.5) = 8.11
Here is the way I see your problem. It isn't a buffer. PLEASE let me know if this is correct.
[HA] = 0.10 and mmols HA in 1L = 100
[A-] = 5.012E-8 and mmols A^- in 1L = 5.012E-5 with
added HCl = 1.5mL x 3.0M = 4.5 mmols.
................A^- + H^+ ==> HA
I........5.012E-5 + 0.............100
add....................4.5.........................
C.........-4.5......-4.5..................+4.5
E...........negative..0.................104.5
So there is no buffering. (H^+) = mmols/mL = 104.5/1000 = 0.1045
pH = -log(H^+) = -log(0.1045) = 0.98
mmols A^- = 0.044 x 1000 = 44
mmols HA = 0.056 x 1000 = 56
mmols HCl added = 1.5 mL x 3.0 M = 4.5, then
..............A^- + H^+ ==> HA
I............44.......0............56
add..................4.5......................
C..........-4.5....-4.5........+4.5
E..........39.5.......0...........60.5
pH = 8.3 + log (39.5/60.5) = 8.11
Here is the way I see your problem. It isn't a buffer. PLEASE let me know if this is correct.
[HA] = 0.10 and mmols HA in 1L = 100
[A-] = 5.012E-8 and mmols A^- in 1L = 5.012E-5 with
added HCl = 1.5mL x 3.0M = 4.5 mmols.
................A^- + H^+ ==> HA
I........5.012E-5 + 0.............100
add....................4.5.........................
C.........-4.5......-4.5..................+4.5
E...........negative..0.................104.5
So there is no buffering. (H^+) = mmols/mL = 104.5/1000 = 0.1045
pH = -log(H^+) = -log(0.1045) = 0.98
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