. An enzyme-catalyzed reaction was carried out in a 0.2M �Tris� buffer, pH 7.8. As a result of the reaction, 0.03 mole/liter of H+ was produced. (pKa of Tris =8.1) (a) What was the ratio of Tris+ (conjugate acid) / Triso (Conjugate base) at the start of the reaction? (b) What are the concentration of Tris+ and Triso at the start of the reaction? (c) Show the reaction by which the buffer maintained a near constant pH. (d) What were the concentration of Tris+ and Triso at the end of the reaction? (e) What was the pH at the end of the reaction? (f) What would the final pH be if no buffer were present?
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Sure, I can help you with that! Here are the step-by-step solutions to your questions:
(a) To find the ratio of Tris+ (conjugate acid) / Triso (conjugate base) at the start of the reaction, we can use the Henderson-Hasselbalch equation:
pH = pKa + log([Tris+]/[Triso])
Rearranging the equation, we can solve for the desired ratio:
[Tris+]/[Triso] = 10^(pH - pKa)
Substituting the given values:
[Tris+]/[Triso] = 10^(7.8 - 8.1) = 10^(-0.3) = 0.501
Therefore, the ratio of Tris+ to Triso at the start of the reaction is approximately 0.501.
(b) To determine the concentrations of Tris+ and Triso at the start of the reaction, we need to convert the ratio into actual concentrations. Let's assume the initial concentration of Triso is x M, then the initial concentration of Tris+ would be 0.501x M.
Given that the buffer concentration is 0.2 M, we can write the equation:
x + 0.501x = 0.2
Simplifying,
1.501x = 0.2
x = 0.133 M
Therefore, the concentration of Triso at the start of the reaction is 0.133 M, and the concentration of Tris+ is 0.501 * 0.133 = 0.0666 M.
(c) The reaction by which the buffer maintains a near constant pH is the ionization of Tris:
Tris+ ⇌ Triso + H+
(d) At the end of the reaction, the amount of H+ produced is given as 0.03 mole/liter. Since H+ is a product of the reaction, it will increase the concentration of Tris+ and decrease the concentration of Triso.
Let's assume the change in concentration of Tris+ is Δ[Tris+], and the change in concentration of Triso is Δ[Triso].
The change in H+ concentration is equal to Δ[Tris+], so Δ[Tris+] = 0.03 M.
Using the stoichiometry of the equation, Δ[Triso] = -Δ[Tris+] = -0.03 M.
Therefore, the concentration of Tris+ at the end of the reaction is 0.0666 M + 0.03 M = 0.0966 M, and the concentration of Triso is 0.133 M - 0.03 M = 0.103 M.
(e) To find the pH at the end of the reaction, we can use the Henderson-Hasselbalch equation again:
pH = pKa + log([Tris+]/[Triso])
Substituting the values at the end of the reaction,
pH = 8.1 + log(0.0966/0.103) = 8.1 - 0.034 = 8.066
Therefore, the pH at the end of the reaction is 8.066.
(f) If no buffer were present, the pH would change significantly with the addition of H+. This is because a buffer helps maintain a constant pH by absorbing H+ ions. Without a buffer, the pH would be determined by the concentration of H+ ions produced. However, without further information about the H+ concentration or any other factors affecting the pH, we cannot determine the final pH precisely.
Sure, I can help you with that! Let's break down each question one by one:
(a) To find the ratio of Tris+ (conjugate acid) to Triso (conjugate base) at the start of the reaction, we can use the Henderson-Hasselbalch equation:
pH = pKa + log([A-]/[HA])
In this case, [A-] represents the concentration of Triso (conjugate base) and [HA] represents the concentration of Tris+ (conjugate acid). We can rearrange the equation to solve for the ratio [A-]/[HA]:
[A-]/[HA] = 10^(pH - pKa)
Given that the pH is 7.8 and the pKa of Tris is 8.1, we can substitute these values into the equation:
[A-]/[HA] = 10^(7.8 - 8.1) = 10^(-0.3) = 0.501
Therefore, the ratio of Tris+ to Triso at the start of the reaction is approximately 0.501.
(b) To find the concentrations of Tris+ and Triso at the start of the reaction, we need to use the information given. We know that the total buffer concentration is 0.2M, and at the start, the ratio of Tris+ to Triso is 0.501.
Let's assume the concentration of Tris+ is x. Then, the concentration of Triso would be 0.501x, as per the ratio.
Since the total buffer concentration is 0.2M, we have the equation:
x + 0.501x = 0.2
Simplifying this equation, we find:
1.501x = 0.2
x = 0.133 M
Therefore, the concentration of Tris+ at the start of the reaction is 0.133 M, and the concentration of Triso is 0.501 times that, which is approximately 0.067 M.
(c) The reaction by which the buffer maintains a near constant pH can be represented as follows:
Tris+ + H2O ⇌ Triso + H3O+
In this reaction, Tris+ acts as a weak acid, donating a proton (H+) to water, resulting in the formation of Triso (conjugate base) and hydronium ion (H3O+). This reaction helps to neutralize any added acid (H+) or base (OH-) and maintain the pH of the buffer.
(d) To find the concentrations of Tris+ and Triso at the end of the reaction, we need to consider the H+ produced during the reaction.
Given that 0.03 mole/liter of H+ was produced, we can assume that the concentration of Triso is increased by this amount, while the concentration of Tris+ is decreased by the same amount. So, the updated concentrations become:
Tris+ = 0.133 M - 0.03 M = 0.103 M
Triso = 0.067 M + 0.03 M = 0.097 M
Therefore, the concentration of Tris+ at the end of the reaction is 0.103 M, and the concentration of Triso is 0.097 M.
(e) To find the pH at the end of the reaction, we can use the Henderson-Hasselbalch equation again:
pH = pKa + log([A-]/[HA])
Using the updated concentrations of Triso (0.097 M) and Tris+ (0.103 M), and the pKa of Tris (8.1), we can substitute these values into the equation:
pH = 8.1 + log(0.097/0.103) ≈ 7.91
Therefore, the pH at the end of the reaction is approximately 7.91.
(f) If no buffer were present, the pH would be affected significantly by the addition of H+ produced during the reaction. Without the buffer, the pH would not remain constant and could show a more significant change.
I hope this helps! Let me know if there's anything more I can assist you with.
a = acid
b = base
pH = pKa + log (b)/(a)
7.8 = 8.1 + log (b/a)
Solve for b/a. This is equation 1 and is the answer for part(a)
equation 2 is a + b = 0.2
Solve these two equations simultaneously to obtain a and b at the beginning. This gives the answer to part (b).
Can you take it from here.