Assignment 1
Question
Consider a monohydrogen phosphate ( HPO42-) and dihydrogen phosphate (H2PO4-) buffer solution.
[HPO42-] = 0.063M
[H2PO4-] = 0.10M
What happens when you add 1.0 ml of 0.10 M HCl to the a 99ml solution?
What would the pH of the solution be without the buffer?
What would the pH of the solution be with the buffer?
Answer
For your answer you will need to go beyond simply filling in the formula and giving me a number.
You will need to convince me that you know what is going on here. Words are required for the explanation not just numbers.
You fill in the words. I will get you started on the numbers.
pH without the buffer? Is that the pH of 1 mL of 0.1M HCl? If so, then
(HCl) = 0.1M and pH = -log(0.1) = 1
pH buffer before addition of HCl.
pH = pK2 + log (base)/(acid). Substitute and solve for pH.
pH with 1 mL 0.1M HCl added.
millimols H2PO4^- = 0.1M x 99 mL = 9.90
millimols HPO4^2- = 0.063 x 99 = 6.24
millimols HCl = 0.1M x 1 mL = 0.1
..........HPO4^- + H^+ ==> H2PO4^2-
I.........6.24.....0........9.90
add...............0.1...........
C........-0.1....-0.1........+0.1
E.........6.14.....0.........10.0
Substitute E line into H-H equation and solve for pH.
Note: Some profs are picky and will count off for using millimoles and not mols/L in the above ICE chart and the following HH equation. That's because the HH equation is ....(base)/(acid) and not mmols/mmols. However, the answer comes out the same BECAUSE concn = mols/L = mmols/mL and since you are using base = mmols/mL and acid = mmols/mL, the mL is ALWAYS the same so they cancel and you are left with mmols/mmols and things work out fine. However, if your prof is picky (I counted off) you have two choices.
1. Divide EACH of the mmols by the total volume (in this case it is 100 mL) and use the concn then in mmols/mL = M or
2. Use mmols as shown in the ICE table above BUT divide each by V (for volume). It would look like this
pH = pK2 + log (10.0/V)/(5.14/V), cancel the Vs and continue with the problem.
Hope this helps. Post any further questions; however, explain in detail what you don't understand.
When 1.0 ml of 0.10 M HCl is added to the 99 ml solution, it will react with the HPO42- and H2PO4- ions in the buffer solution. HCl is a strong acid, so it will completely dissociate in water, releasing H+ ions. The H+ ions will combine with the phosphate ions in the buffer solution to form H2PO4- (dihydrogen phosphate). This reaction can be represented as follows:
HPO42- + H+ → H2PO4-
Since the concentration of H2PO4- in the buffer solution is already 0.10 M, the addition of HCl will increase its concentration. This increase in H2PO4- concentration will lead to a decrease in OH- concentration since the reaction consumes H+ ions. As a result, the solution will become more acidic.
Without the buffer, the pH of the solution can be calculated using the Henderson-Hasselbalch equation:
pH = pKa + log ([A-]/[HA])
where pKa is the acid dissociation constant of the acid/base pair, [A-] is the concentration of the conjugate base, and [HA] is the concentration of the acid.
In this case, the acid/base pair is H2PO4-/HPO42- and their pKa is given as 7.20. Without the buffer, the concentration of H2PO4- will increase due to the addition of HCl, while the concentration of HPO42- will remain constant. Substituting the values into the Henderson-Hasselbalch equation, the pH of the solution without the buffer can be calculated.
However, without knowing the concentration of H3PO4 (phosphoric acid), it is not possible to accurately calculate the pH of the solution with the buffer. The buffer capacity depends on the ratio of the concentrations of the conjugate base (HPO42-) to the acid (H2PO4-). So, without the concentration of H3PO4, it is not possible to accurately determine the pH of the solution with the buffer.
To understand what happens when you add 1.0 ml of 0.10 M HCl to the solution, we need to consider the reaction that occurs between HCl and the components of the buffer solution. HCl is a strong acid, meaning it completely dissociates in water to produce hydrogen ions (H+).
When HCl is added, it reacts with the monohydrogen phosphate (HPO42-) and dihydrogen phosphate (H2PO4-) ions in the buffer solution. The H+ ions from HCl react with the HPO42- ions in the buffer, resulting in the formation of H2PO4- ions. This reaction is described by the equation:
HPO42- + H+ ⇌ H2PO4-
The equilibrium shifts to the right to compensate for the increase in H+ ions from HCl. As a result, some of the HPO42- ions are converted to H2PO4- ions.
Next, let's consider the pH of the solution without the buffer. Since HCl is a strong acid, it completely dissociates to produce H+ ions and Cl- ions. The pH of the solution would be determined solely by the concentration of H+ ions, which is determined by the concentration of HCl. Therefore, the pH of the solution without the buffer would be determined by the concentration of HCl, assuming no other acid-base reactions occur.
On the other hand, when a buffer is present, it helps resist changes in pH. In this case, the buffer components, HPO42- and H2PO4-, are a conjugate acid-base pair. The HPO42- ions act as a weak base and the H2PO4- ions act as a weak acid.
When the HCl is added to the buffer solution, it reacts with the HPO42- ions to form H2PO4- ions. However, the buffer system can counteract this change in pH because it can absorb or donate H+ ions without significantly affecting the overall pH.
The buffer acts by shifting the equilibrium between HPO42- and H2PO4- to maintain a relatively constant pH. As H+ ions are added from the HCl, they react with some of the HPO42- ions to form H2PO4- ions. This helps to maintain the equilibrium and prevent a drastic change in pH.
To calculate the pH of the solution with the buffer, we need to use the Henderson-Hasselbalch equation:
pH = pKa + log ([A-]/[HA])
Where pKa is the dissociation constant of the weak acid (H2PO4-), [A-] is the concentration of the conjugate base (HPO42-), and [HA] is the concentration of the weak acid (H2PO4-).
In this case, the pKa can be determined from the dissociation constant of phosphoric acid (H3PO4), as it is the parent compound of the buffer components.
By substituting the given concentrations of [HPO42-] and [H2PO4-] into the equation, we can calculate the pH of the buffer solution.