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?
Sara, do you have answer choices? The answer I'm getting doesn't seem realistic so I may be missing something.
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.
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
To find the pH after adding HCl to the Tris buffer, we need to consider the effect of dilution and the reaction between the added acid and the buffer components.
First, let's analyze the reaction between HCl and the buffer components. Tris (often referred to as THAM) is a weak base that can act as both an acid (HA) and its conjugate base (A-).
The dissociation reaction of Tris buffer can be represented as follows:
HA ⇌ H+ + A-
The equilibrium constant expression for this reaction is given by:
Ka = [H+][A-] / [HA]
Given that the pKa of Tris is 8.3, we can use this information to calculate the initial concentrations of HA and A- in the buffer before adding HCl.
For a buffer at pH 2.0, we know that the concentration of H+ ions is 10^(-pH) = 0.01 M. Using this information and the equilibrium constant expression, we can determine the concentrations of HA and A- in the original buffer.
[HA]/[A-] = [H+]/Ka = 0.01 / (10^(-pKa))
Substituting the values, we find:
[HA] = 0.01 / (10^(8.3)) = 0.0595 M
[A-] = 0.01 - 0.0595 = 0.0005 M
Now, let's consider the dilution. When 1.5 mL of 3.0 M HCl is added to 1.0 L of the buffer, the total volume of the resulting solution becomes 1.0 L + 0.0015 L = 1.0015 L.
To determine the new concentration of H+ ions, we should calculate the moles of H+ ions added by the HCl solution and the volume of the resulting solution.
Moles of HCl added = 1.5 mL * (3.0 mol/L) * (1 L / 1000 mL) = 0.0045 mol
Now, we need to calculate the final concentration of H+ ions in the solution. Since the total volume has increased, the concentration will decrease due to dilution.
To find the new concentration of H+ ions, divide the moles of H+ ions by the total volume in liters:
[H+] = (0.0045 mol) / 1.0015 L = 0.004497 M
Finally, we can calculate the new pH using the equation pH = -log([H+]):
pH = -log(0.004497) = 2.35
Therefore, the pH after adding 1.5 mL of 3.0 M HCl to 1.0 L of the Tris buffer is approximately 2.35.