what is the pH of the buffer when 4ml of 1M HCl is added to a 100ml buffer solution made using 5ml 1M Na2HPO4 and 5ml 1M NaH2PO4? The pKa for H+ + HPO4- ==> H2PO4 is 6.82.

What would happen to the pH if the 100ml buffer had been made up at the starting pH (6.82) using Tris (total Tris concentration is 100mM) instead? Tris has a pKa of 8.08.

To determine the pH of the buffer solution after adding 4 ml of 1M HCl, we need to consider the Henderson-Hasselbalch equation. The Henderson-Hasselbalch equation is given by:

pH = pKa + log ([A-] / [HA])

Where:
pH = the pH of the buffer solution
pKa = the dissociation constant of the weak acid (H2PO4- in this case)
[A-] = concentration of the conjugate base (HPO4- in this case)
[HA] = concentration of the weak acid (H2PO4- in this case)

Step 1: Calculate the concentration of HPO4- and H2PO4- in the buffer.
The initial concentration of H2PO4- and HPO4- are the same because both Na2HPO4 and NaH2PO4 are added in equal amounts.
Concentration of H2PO4- = (5 ml / 100 ml) * 1 M = 0.05 M
Concentration of HPO4- = (5 ml / 100 ml) * 1 M = 0.05 M

Step 2: Determine the moles of H2PO4- and HPO4- in the buffer.
Moles of H2PO4- = 0.05 M * 0.1 L = 0.005 moles
Moles of HPO4- = 0.05 M * 0.1 L = 0.005 moles

Step 3: Calculate the moles of H+ added from HCl.
Moles of HCl = 1 M * 4 ml / 1000 = 0.004 moles

Step 4: Calculate the total moles of H2PO4- and HPO4- after the addition of HCl.
H2PO4- will react with HCl to form H3PO4, and HPO4- will react with H+ to form H2PO4-. So, the moles of H2PO4- will increase, and the moles of HPO4- will decrease.
Moles of H2PO4- = 0.005 moles + 0.004 moles = 0.009 moles
Moles of HPO4- = 0.005 moles - 0.004 moles = 0.001 moles

Step 5: Calculate the new concentrations of H2PO4- and HPO4- in the buffer.
Concentration of H2PO4- = 0.009 moles / 0.1 L = 0.09 M
Concentration of HPO4- = 0.001 moles / 0.1 L = 0.01 M

Step 6: Calculate the pH using the Henderson-Hasselbalch equation.
pH = pKa + log ([A-] / [HA])
pH = 6.82 + log (0.01 M / 0.09 M)
pH ≈ 6.82 + log (0.111)
pH ≈ 6.82 - 0.949
pH ≈ 5.871

Therefore, the pH of the buffer solution after adding 4 ml of 1M HCl is approximately 5.871.

For the second part of the question, we need to consider the effect of using Tris as the buffer instead of Na2HPO4 and NaH2PO4. Tris (tris(hydroxymethyl)aminomethane) acts as a weak base in the buffer system.

When Tris is used as the buffer, the pKa value of Tris (8.08) should be used in the Henderson-Hasselbalch equation instead of the pKa value of H2PO4- (6.82).

Using the same steps as before, we can calculate the pH of the Tris buffer after adding 4 ml of 1M HCl.

pH = pKa + log ([A-] / [HA])
pH = 8.08 + log ([Tris+] / [TrisH])

As the buffer was originally made at the starting pH of 6.82, the initial concentrations of [Tris+] and [TrisH] are the same: 100 mM each.

After the addition of HCl, the moles of TrisH will increase, and the moles of Tris+ will decrease. However, the concentration of both TrisH and Tris+ will remain the same (100 mM) because Tris is a strong enough base to neutralize the added HCl.

Therefore, the pH of the buffer after adding HCl will remain the same as the initial pH of 6.82 when Tris is used as the buffer.