What happens when 4ml of 1M HCL is added to 100ml buffer made using 5ml 1M Na2HPO4 and 5ml 1M NaH2PO4? The pKa for H+ + HPO4- = H2PO4 is 6.82.
Use the Henderson-Hasselbalch equation.
pH = pKa + log [(base)/(acid)]
H^+ + HPO4^-2 ==> H2PO4^-
You start with 0 millimoles H^+.
You start with 5 x 1 = 5 millimoles HPO4^-2
You start with 5 x 1 = 5 millimoles H2PO4^-1.
Now you add 4 x 1 = 4 millimoles H^+.
That reacts with 4 millimiles HPO4^-2 to form 4 millimoles H2PO4^- (plus the 5 already there = 9 millimoles H2PO4^-).
That leaves 5 millimoles HPO4^-2 - 4 millimoles that reacted with H^+ = 1 millimole HPO4^-.
Plug into HH equation and solve for pH.
Check my work. It's easy to leave type a wrong charge, etc.
To determine what happens when 4 ml of 1M HCl is added to a buffer made using 5 ml of 1M Na2HPO4 and 5 ml of 1M NaH2PO4, we need to consider the equilibrium reaction that occurs between the acid and its conjugate base components of the buffer.
In this case, the conjugate acid is H2PO4- and the conjugate base is HPO4^2-. The equilibrium reaction is as follows:
H+ + HPO4^2- ⇄ H2PO4-
The pKa value given, 6.82, is the negative logarithm of the acid dissociation constant (Ka) for this reaction. It represents the acidity strength of the system. A lower pKa value indicates a stronger acid, while a higher pKa value indicates a weaker acid.
Now, let's analyze the addition of 4 ml of 1M HCl to the buffer solution:
Step 1: Calculate the initial concentrations of the components in the buffer:
- Initial concentration of H2PO4- = (5 ml / 100 ml) * 1 M = 0.05 M
- Initial concentration of HPO4^2- = (5 ml / 100 ml) * 1 M = 0.05 M
Step 2: Calculate the final concentrations after the addition of HCl:
- The moles of H+ added = (4 ml / 1000 ml) * 1 M = 0.004 moles
- The moles of H2PO4- in the buffer = (0.05 M) * (0.01 L) = 0.0005 moles
- The moles of HPO4^2- in the buffer = (0.05 M) * (0.01 L) = 0.0005 moles
Step 3: Use the Henderson-Hasselbalch equation to determine the pH of the buffer solution after the addition of HCl:
- pH = pKa + log10([HPO4^2-] / [H2PO4-])
- pH = 6.82 + log10(0.0005 / 0.0005)
- pH = 6.82
So, after adding 4 ml of 1M HCl to the buffer solution, the pH remains unchanged at 6.82. This is because the acid and base components in the buffer effectively neutralize the added HCl while maintaining the original pH of the system.