Calculate pH when 0.060 mols of HCl are added to 1.000 L of a buffer containing 0.499M KH2PO4 and 0.193M K2HPO4
Use the Henderson-Hasselbalch equation and the ICE chart below.
moles HPO4^2- = 0.193
moles H2PO4^- = 0.499
...........HPO4^2- + H^+ ==> H2PO4^-
initial....0.193......0.......0.499
add..................0.06...............chchange.....-0.06.....0.06......+0.06
equil.......0.133......0........0.493
To calculate the pH of the buffer solution after adding HCl, we need to consider the dissociation of the acids and their conjugate bases.
The components of the buffer solution are KH2PO4 (acid) and K2HPO4 (conjugate base). When a weak acid and its conjugate base are combined, a buffer solution is formed that resists changes in pH.
First, we need to determine the initial concentrations of the acid and conjugate base before the addition of HCl.
Given:
- Initial concentration of KH2PO4 (acid) = 0.499 M
- Initial concentration of K2HPO4 (conjugate base) = 0.193 M
Since KH2PO4 is a weak acid, it will partially dissociate in water as follows:
KH2PO4 ⇌ H+ + H2PO4-
The amount of H+ produced by the dissociation of KH2PO4 will be equal to the amount of H2PO4- present initially.
So, we have:
[H+] = [H2PO4-] = 0.499 M
Now, let's calculate the pH of the buffer solution after the addition of 0.060 moles of HCl.
HCl is a strong acid, so it will completely dissociate in water to produce H+ ions.
The final volume of the solution after adding HCl is 1.000 L.
To determine the final concentration of H+ ions, we need to consider the moles of H+ ions from HCl and those contributed by the dissociation of H2PO4-.
The moles of H+ ions contributed by HCl = 0.060 mol
The total moles of H+ ions = moles from HCl + moles from H2PO4- = 0.060 mol + 0.499 mol (initial [H2PO4-] concentration) = 0.559 mol
The final concentration of H+ ions = moles / volume = 0.559 mol / 1.000 L = 0.559 M
To calculate the pH, we can use the formula:
pH = -log[H+]
pH = -log(0.559)
pH ≈ 0.253
Therefore, the pH of the buffer solution after adding 0.060 moles of HCl is approximately 0.253.