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.
I answered this earlier.
Use the Henderson-Hasselbalch equation.
To determine the pH of the buffer mixture, we need to calculate the concentration of the protonated form (H2PO4-) and the disassociated form (HPO4^2-) of the phosphate buffer system. Given that the pKa of H3PO4 is 2.1, we can use the Henderson-Hasselbalch equation:
pH = pKa + log ([A-]/[HA])
First, let's calculate the amount of the protonated form (H2PO4-) in the mixture:
Using the formula C1V1 = C2V2, where C represents concentration and V represents volume:
Initial moles of NaH2PO4 = 0.1 mol/L × 0.4 L = 0.04 mol
Final moles of NaH2PO4 = 0.04 mol + 0.01 mol = 0.05 mol
Now, let's calculate the amount of the disassociated form (HPO4^2-) in the mixture:
Initial moles of H3PO4 = 0.1 mol/L × 0.1 L = 0.01 mol
Final moles of H3PO4 = 0.04 mol + 0.01 mol = 0.05 mol
Now, let's calculate the concentrations of H2PO4- and HPO4^2- in the buffer mixture:
[H2PO4-] = moles H2PO4- / total volume
[HPO4^2-] = moles HPO4^2- / total volume
[H2PO4-] = 0.05 mol / (0.4 L + 0.1 L) = 0.1 M
[HPO4^2-] = 0.05 mol / (0.4 L + 0.1 L) = 0.1 M
Now, let's plug these values into the Henderson-Hasselbalch equation:
pH = 2.1 + log ([0.1 M]/[0.1 M])
pH = 2.1 + log (1)
pH = 2.1
Therefore, the pH of the buffer mixture is 2.1.