If 10mL of 1 M HCl is added to 1L of 0.01 M phosphate buffer, pH 7.2, what is the resulting pH?
Also what are the conc of H2PO4^1- and HPO4^2- in the final solution?
HCl:
M = mol / L
1M = mol / 0.01L
= 0.01 mol
phosphate:
M = mol / L
0.01M = mol / 1L
= 0.01 mol
H2PO4 = 0.005mol
HPO4 = 0.005mol
so im lost here....
using HH equation
pH= 2.14 +log(0.005)/(0.005)
pH = 2.14
pH= 2.14 from question
thanks a lot
I think the pH of 2.14 is correct but this is really a complex question. As for the concentrations, H2PO4^- = 5 mmols/1010 mL = ?
HPO4^2- = 0
H3PO4 is 5 mmols/1010 mL but the problem doesn't ask for that.
I approached the problem this way. The HH equation with the pH of 7.20 is
7.20 = 7.20 + log (b/a) so b/a = 1
Then you know a + b = 0.01 x 1000 = 10 mmols and solving the two equations simultaneously, even though you already know it, a = b in mmols which means a = b = 5 mmols.. So
...............HPO4^2- + HCl ==> H2PO4^- + ...
I................5.................0...............5
add.............................10....................
C............-5.................-5...............+5
E............. 0..................5................10
It's important here to realize that there is enough HCl added to convert ALL of the HPO4^2- to H2PO4^- and have an excess of 5 mmols HCl remain which can then react with H2PO4^- to form H3PO4. You put those new values into a new HH equation, but this time with k1 and not k2. I hope this helps.
To find the resulting pH, we can use the Henderson-Hasselbalch equation, which is commonly used to calculate the pH of a buffer solution. The equation is as follows:
pH = pKa + log([A-]/[HA])
In this case, the buffer system is H2PO4^-/HPO4^2-, and the pKa of this system is approximately 7.2. The concentrations of the two components of the buffer, [H2PO4^-] and [HPO4^2-], are equal to 0.01 M each.
Now, when 10 mL of 1 M HCl is added to 1 L of the buffer solution, the HCl will react with the H2PO4^- in the buffer to form water and H3PO4. The amount of HCl added is 0.01 mol.
The moles of H2PO4^- initially present in the buffer solution is 0.01 mol, and the moles of HCl added is also 0.01 mol. So, after the reaction, the resulting concentration of H2PO4^- will be 0.01 - 0.01 = 0 M.
As for HPO4^2-, it does not react with HCl, so the initial concentration of HPO4^2- remains the same at 0.01 M.
Now, we can calculate the resulting pH using the Henderson-Hasselbalch equation:
pH = 7.2 + log(0/0.01)
Since log(0) is undefined, we have an issue here. When all the H2PO4^- is consumed during the reaction, the HPO4^2- starts acting as a base and reacts with the HCl. As a result, the solution will become acidic.
Therefore, the resulting pH cannot be determined solely based on the information provided. However, we can conclude that the pH will be lower than the initial pH of 7.2.
As for the concentration of H2PO4^- and HPO4^2- in the final solution, the concentration of H2PO4^- will be 0 M, as explained earlier. The concentration of HPO4^2- remains the same at 0.01 M.