an ezyme -catalysed reaction was carried out in a solution buffered with 0.05 M phosphate ph 7.2 . As a result of the reaction ,0.06M of acid was formed. (phosphoric acid has three pka values. The one required for this calculation is
pka2=7.2).
1) what was the pH at the end of the reaction .
2) write the chemical equation showing how the phosphate buffer resisted a large charge in pH.
Josh, did you proof your problem? You can calculate the original (acid) and (base) this way.
pH = pKa2 + log base/acid
7.2 = 7.2 + log base/acid
0 = log b/a
1 = b/a is equation 1
equation 2 is a + b = 0.05
Solve those two equations simultaneously and A = B = 0.025M
The buffer reaction is
HPO4^2- + H^+ ==> H2PO4^-
The final pH, of course, depends upon the H+ added but 0.06 M sounds unrealistic to me. Could that be 0.006M?
To answer the questions, we need to understand the concept of a buffer and how it works. A buffer is a solution that resists changes in pH when small amounts of acid or base are added. It consists of a weak acid and its conjugate base (or vice versa) in roughly equal amounts.
Let's start with the first question:
1) What was the pH at the end of the reaction?
In the given scenario, the enzyme-catalyzed reaction resulted in the formation of acid. Since we know the initial pH of the buffer solution and the concentration of the formed acid, we can use the Henderson-Hasselbalch equation to calculate the pH change.
The Henderson-Hasselbalch equation is given as:
pH = pKa + log ([A-]/[HA])
In this equation:
- pH is the final pH we are trying to find.
- pKa is the logarithmic form of the acid dissociation constant for the weak acid in the buffer. In this case, the required pKa value is given as pKa2 = 7.2.
- [A-] is the concentration of the conjugate base (in this case, the concentration of phosphate ions).
- [HA] is the concentration of the weak acid (in this case, the concentration of the acid formed, which is 0.06 M).
Using the provided values, we can substitute them into the Henderson-Hasselbalch equation and solve for the final pH:
pH = 7.2 + log ([A-]/[HA])
Before we can proceed, we need to determine the concentrations of [A-] and [HA].
Since the reaction resulted in the formation of acid, the stoichiometry of the reaction suggests that the concentration of [A-] is the same as the initial concentration of the initially buffered [HA], which is 0.05 M.
So, substituting the values into the equation:
pH = 7.2 + log (0.05/0.06)
Using logarithmic properties and calculating the division, we get:
pH = 7.2 - 0.0791
Therefore, the pH at the end of the reaction is approximately 7.1209, or simply 7.1 (rounded to one decimal place).
2) Write the chemical equation showing how the phosphate buffer resisted a large change in pH.
The phosphate buffer, in this case, consists of the weak acid H2PO4- (dihydrogen phosphate ion) and its conjugate base HPO42- (monohydrogen phosphate ion).
The chemical equation representing the buffer action can be written as follows:
H2PO4- ⇌ H+ + HPO42-
When a small amount of acid is formed during the reaction, the excess H+ ions are consumed by the phosphate buffer, shifting the equilibrium to the left. Similarly, if a small amount of base is added, the excess OH- ions react with H+ ions, shifting the equilibrium to the right.
This buffering action helps maintain the pH of the solution relatively stable, as the concentrations of the acidic species (H2PO4-) and the basic species (HPO42-) can adjust to stabilize the pH.