if 400 mL of 0.1 M NaH2PO4 is mixed with 100 mL of 0.1 M H3PO4 what is the pH of the buffer mixture? The PKa of H3PO4 is 2.1.
Use the Henderson-Hasselbalch equation.
pH = 2.1 + log(40/10) = ??
To calculate the pH of a buffer mixture, we need to determine the concentrations of the acid (H3PO4) and its conjugate base (H2PO4^-) in the solution.
First, let's calculate the moles of H3PO4 and NaH2PO4 in each solution:
For 400 mL of 0.1 M NaH2PO4:
Number of moles of NaH2PO4 = concentration (M) x volume (L)
= 0.1 M x 0.4 L
= 0.04 moles
For 100 mL of 0.1 M H3PO4:
Number of moles of H3PO4 = concentration (M) x volume (L)
= 0.1 M x 0.1 L
= 0.01 moles
Next, let's determine the concentrations of H3PO4 and H2PO4^- in the buffer solution.
H3PO4 is a weak acid and partially dissociates into H2PO4^- and H+ ions. Using the dissociation equation:
H3PO4 ⇌ H2PO4^- + H+
For NaH2PO4, the initial number of moles of H2PO4^- is equal to the number of moles of NaH2PO4, which is 0.04 moles. However, initially, there is no H3PO4, so the initial number of moles of H3PO4 is 0.
At equilibrium, the number of moles of H3PO4 formed is equal to the number of moles of NaH2PO4 dissociated, which is also 0.04 moles. The number of moles of H2PO4^- formed is equal to the number of moles of NaH2PO4 minus the number of moles of H3PO4 formed, which is 0.04 - 0 = 0.04 moles.
The total volume of the buffer solution is 400 mL + 100 mL = 500 mL = 0.5 L.
Now, we can calculate the concentrations:
[H3PO4] = moles / volume
= 0.01 moles / 0.5 L
= 0.02 M
[H2PO4^-] = moles / volume
= 0.04 moles / 0.5 L
= 0.08 M
Since we know the pKa of H3PO4 is 2.1, we can use the Henderson-Hasselbalch equation to calculate the pH of the buffer solution.
pH = pKa + log([H2PO4^-] / [H3PO4])
= 2.1 + log(0.08 M / 0.02 M)
= 2.1 + log(4)
≈ 2.1 + 0.602
≈ 2.702
Therefore, the pH of the buffer mixture is approximately 2.702.
To find the pH of the buffer mixture, we need to consider the acid-base equilibrium between NaH2PO4 and H3PO4.
The reaction between NaH2PO4 and H3PO4 can be written as follows:
NaH2PO4 + H3PO4 ⇌ Na2HPO4 + H2O
Since NaH2PO4 and H3PO4 are both weak acids, we can use the Henderson-Hasselbalch equation to calculate the pH of the buffer mixture.
The Henderson-Hasselbalch equation is given by:
pH = pKa + log ([A-]/[HA])
In this case, [A-] represents the concentration (in moles per liter) of the conjugate base (Na2HPO4), and [HA] represents the concentration of the acid (H3PO4).
We are given that the pKa of H3PO4 is 2.1. To calculate the concentrations of [A-] and [HA], we need to take into account the volume and concentration of the solutions.
First, let's calculate the number of moles of NaH2PO4 and H3PO4 in the solutions.
moles of NaH2PO4 = (volume of NaH2PO4 × molarity of NaH2PO4) / 1000
moles of NaH2PO4 = (400 mL × 0.1 M) / 1000 = 0.04 moles
moles of H3PO4 = (volume of H3PO4 × molarity of H3PO4) / 1000
moles of H3PO4 = (100 mL × 0.1 M) / 1000 = 0.01 moles
Now, we can calculate the concentrations:
[A-] = moles of Na2HPO4 / total volume of the buffer mixture
[A-] = (0.04 moles) / (400 + 100) mL = 0.04 M
[HA] = moles of H3PO4 / total volume of the buffer mixture
[HA] = (0.01 moles) / (400 + 100) mL = 0.01 M
Now, we can substitute the values into the Henderson-Hasselbalch equation to find the pH:
pH = 2.1 + log (0.04/0.01)
pH = 2.1 + log (4)
pH = 2.1 + 0.6
pH ≈ 2.7
Therefore, the pH of the buffer mixture is approximately 2.7.