An enzyme-catalyzed reaction is carried out in a 50-mL solution containing 0.1 M TRIS buffer. The pH of the reaction mixture at the start was 8.0. As a result of the reaction, 0.002 mol of H+ were produced. What is the ratio of TRIS base to TRIS acid at the start of the experiment? What is the final pH?
If needed: TRIS(Trizma base) mw= 121.1; pka= 8.3; TRIS-HCl mw= 157.6; pka= 8.3
I kind of have an idea. I start out with the hendersn hasselbalch equation and solve for the ratio ([Tris] / [Tris HCl]). I need help doing this first before i can proceed with the rest of the problem.
Thank you very much.
To find the ratio of TRIS base to TRIS acid at the start of the experiment, you can use the Henderson-Hasselbalch equation. The equation is given by:
pH = pKa + log([A-]/[HA])
Where pH is the initial pH (8.0), pKa is the dissociation constant (8.3), and [A-] and [HA] are the concentrations of TRIS base and TRIS acid, respectively.
Rearranging the equation, we can solve for the ratio:
[A-]/[HA] = 10^(pH - pKa)
Substituting the values, we get:
[A-]/[HA] = 10^(8.0 - 8.3)
[A-]/[HA] = 10^(-0.3)
[A-]/[HA] = 0.501
So, the ratio of TRIS base to TRIS acid at the start of the experiment is approximately 0.501.
To find the final pH after the production of H+, we need to consider the stoichiometry of the reaction. Given that 0.002 mol of H+ were produced, we need to determine how this affects the pH.
Since TRIS is acting as a buffer, it can either accept or donate H+ ions to maintain the pH. The TRIS base (A-) will accept H+ ions, while the TRIS acid (HA) will donate H+ ions.
Let's assume x moles of TRIS base (A-) are converted to TRIS acid (HA). Then, we can write the equation representing this reaction:
x = 0.002 mol
The remaining concentrations of TRIS base and TRIS acid can be calculated:
[HA] = [HA]_initial + x mol
[A-] = [A-]_initial - x mol
Since the volumes are equal, the final concentrations can be expressed as:
[HA] = [HA]_initial + x mol/50 mL
[A-] = [A-]_initial - x mol/50 mL
Now, we can calculate the final pH using the Henderson-Hasselbalch equation:
pH = pKa + log([A-]/[HA])
Substituting the values, we get:
pH = 8.3 + log(([A-]_initial - x mol/50 mL) / ([HA]_initial + x mol/50 mL))
Simplifying the equation, we can substitute the ratio of [A-]/[HA] at the start of the experiment (0.501):
pH = 8.3 + log(0.501 - x/50 mL / (0.001 + x/50 mL))
Solving for x, we can use the given information that x = 0.002 mol:
pH = 8.3 + log(0.501 - 0.002/50 mL / (0.001 + 0.002/50 mL))
Calculating this equation will give you the final pH.