Question

In: Chemistry

You want to maintain pH = 7.0 for an enzyme-catalyzed reaction that will produce hydrogen ions...

You want to maintain pH = 7.0 for an enzyme-catalyzed reaction that will produce hydrogen ions along with the desired product. At equal concentrations, which weak acid, if any, will serve as the better buffer for the reaction: Acid A pKa = 6.5, Acid B pKa = 7.5

a. Acid A

b. both are equally effective

c. water is as good as either of the acids available

d. acid B

*** I get that answer A is correct but Wouldn't B (pKa = 7.5) be better because as more H+ gets released by the reaction the pH would go down closer to 7, while if A was used the pH would fall further away from 7 . Also I'm not sure what is at "equal concentrations" as in the question

Solutions

Expert Solution

The correct answer is A. Weak acid A (pKa = 6.5).

- If we are working with a buffer, it means that we have a weak acid or base with its conjugated salt (conjugated basic or acid salt). In this case we have a weak acid with its conjugated basic salt.

- At equal concentrations: it means that you have a concentration of 0.1 M of Acid A and 0.1 M of Acid B.

So, which to choose and why?

Lets put an example. Here I am going to use the HH equation (Henderson-Hasselbach). This equation is used with Buffer solutions or common ion effect solutions.

Henderson-Hasselbach equation: pH = pKa + log ( base / acid )

1 -----> Let's say that we have 0.1 M of Acid A with 0.2 M of its conjugated basic salt. pKa = 6.5.

pH = pKa + log [base (conjugated salt) / acid (Acid A)]

pH = 6.5 + log ( 0.2 / 0.1 )

pH = 6.5 + 0.30

pH = 6.8

2 -------> Let's say that we have 0.1 M of Acid B with 0.2 M of its conjugated basic salt. pKa = 7.5

pH = pKa + log [base (conjugated salt) / acid (Acid B)]

pH = 7.5 + log ( 0.2 / 0.1 )

pH = 7.5 + 0.30

pH = 7.8

According to calculations, we see that with Acid A we obtain a pH close to 7, while with Acid B we obtain a pH > than 7.

So, if we choose a weak acid with a higher pKa (Acid B) the pH is going to be higher than 7.

If we choose a weak acid with a lower pKa (Acid A) the pH is going to maintain close to 7.

If you need more explanation or if you have any doubt please you could leave a comment.

queen_honey_blossom answered 1 year ago
Next > < Previous

Related Solutions

Question 2-17 You want to maintain a pH = 7.0 for an enzyme-catalyzed reaction that will...
Question 2-17 You want to maintain a pH = 7.0 for an enzyme-catalyzed reaction that will produce hydrogen ions along with the desired product. At equal concentrations, which weak acid, if any, will serve as the better buffer for the reaction: Acid A, with pKa = 6.5 or Acid B, with pKa = 7.5? a. Acid A b. Water is as good as either of the acid available c. Acid B
To the nearest hundredths, what is the pH at the end of an enzyme-catalyzed reaction if...
To the nearest hundredths, what is the pH at the end of an enzyme-catalyzed reaction if it were carried out in a 0.1 M buffer initially at pH 6.65, and 0.006 M of acid (H+) was produced during the reaction? (The buffer used was a polyprotic acid of the general form: H3A; the three pKas of this polyprotic acid are 2.11, 6.65, and 11.33.)
To the nearest hundredths, what is the pH at the end of an enzyme-catalyzed reaction if...
To the nearest hundredths, what is the pH at the end of an enzyme-catalyzed reaction if it were carried out in a 0.1 M buffer initially at pH 6.73, and 0.004 M of acid (H+) was produced during the reaction? (The buffer used was a polyprotic acid of the general form: H3A; the three pKas of this polyprotic acid are 2.17, 6.73, and 11.27.)
To the nearest hundredths, what is the pH at the end of an enzyme-catalyzed reaction if...
To the nearest hundredths, what is the pH at the end of an enzyme-catalyzed reaction if it were carried out in a 0.1 M buffer initially at pH 6.73, and 0.004 M of acid (H+) was produced during the reaction? (The buffer used was a polyprotic acid of the general form: H3A; the three pKas of this polyprotic acid are 2.17, 6.73, and 11.27.)
Imagine that you are observing an enzyme-catalyzed reaction in the laboratory. The reaction is progressing as...
Imagine that you are observing an enzyme-catalyzed reaction in the laboratory. The reaction is progressing as expected. As you periodically add more enzyme, the reaction increases proportionally until suddenly it stops increasing. At this point, no matter how much more enzyme that you add, the reaction rate does not change. Assuming no other chemicals have been added or changed, explain why the reaction rate has leveled off.
An enzyme-catalyzed reaction was carried out in a solution buffered with 0.03 M phosphate, PH 7.2....
An enzyme-catalyzed reaction was carried out in a solution buffered with 0.03 M phosphate, PH 7.2. As a result of the reaction, 0.004mole/liter of acid was formed a) What is the pH at the end of the reaction.? b) what would the pH if No buffer were present? Answers for questions a and b are pH = 6.9 and pH =2.4 respectively. BUT HOW DID THEY GET THOSE pH?
25) Am enzyme-catalyzed reaction proceeds at a faster rate than an uncatalyzed reaction because the enzyme...
25) Am enzyme-catalyzed reaction proceeds at a faster rate than an uncatalyzed reaction because the enzyme lowers the difference in free energy between the reactants and the products. a. True b. False
A sample of water has the following concentration of ions (and pH = 7.0).                         cations...
A sample of water has the following concentration of ions (and pH = 7.0).                         cations                                    mg/L      anions            mg/L                         Ca+2      95.0                           HCO3-   160.0                         Mg+2     26.0                           SO42-     135.0                         Na+       15.0                           Cl-         73.0 a) Draw an ion concentration bar (use meq/L for units). b) What is the total hardness (TH) in mg/L as CaCO3? c) What is the carbonate hardness (CH) in mg/L as CaCO3? d) What is the non-carbonate hardness (NCH) in mg/L as CaCO3? e)...
The following reaction rate data are collected upon analysis of an enzyme catalyzed hydrolysis. The reaction...
The following reaction rate data are collected upon analysis of an enzyme catalyzed hydrolysis. The reaction is carried out with two different inhibitors. Determine the modes of inhibition for each inhibitor. (The rates provided are relative to the rate measured in the absence of inhibitor. The enzyme concentration for each set of data is equal.) No inhibitor [S]0 (10−2 M) 1.25 3.84 5.81 7.13 v0 0.398 0.669 0.859 1.000 [I1]– –2mmol/L [S]0 (10−2 M) 1.25 2.5 4.00 5.50 v0 0.179...
Consider the following data for an enzyme catalyzed hydrolysis reaction in the presence and absence of...
Consider the following data for an enzyme catalyzed hydrolysis reaction in the presence and absence of inhibitor I: [Substrate] M                                    vo (µmol/min)                    voI (µmol/min) 6x10-6 20.8 4.2 1x10-5                                                    29                                                           5.8 2x10-5                                                    45                                                           9 6x10-5                                                    67.6                                                        13.6 1.8x10-4                                                87                                                           16.2   Use the above data, do the following: a. Generate Lineweaver-Burk plots of the data. b. Explain the significance of the x-interecpt, y-intercept, and the slope. c.   Identify the type of inhibition.
ADVERTISEMENT
Subjects
ADVERTISEMENT
Latest Questions
ADVERTISEMENT