Calculate the pH of the buffer formed by mixing 50.0mL of 0.200 F NaH2PO4 with 50.0mL of 0.120 F HCl
To calculate the pH of a buffer, we need to consider the equilibrium reaction that takes place between the weak acid and its conjugate base. In this case, NaH2PO4 is a weak acid and its conjugate base is H2PO4-. The equilibrium reaction is as follows:
NaH2PO4 ⇌ Na+ + H2PO4-
The second component of the buffer is HCl, which is a strong acid. However, since the concentration of HCl is lower than that of NaH2PO4, we can assume that most of the added HCl will react with NaH2PO4 to form H2PO4-. This reaction can be represented as follows:
NaH2PO4 + HCl ⇌ Na+ + H2PO4- + H2O
To determine the pH of the buffer, we can use the Henderson-Hasselbalch equation:
pH = pKa + log([A-]/[HA])
Where:
- pH is the measure of acidity or basicity on a logarithmic scale.
- pKa is the negative logarithm of the acid dissociation constant (Ka) of the weak acid.
- [A-] is the concentration of the conjugate base (H2PO4-).
- [HA] is the concentration of the weak acid (NaH2PO4).
First, we need to find the values for pKa, [A-], and [HA].
The pKa value of H2PO4- can be found in reference sources or calculated based on experimental data. For simplicity, let's assume the pKa of H2PO4- is 7.21.
Next, we need to calculate the molarities of [A-] and [HA]. Since we are given the volume and concentration of the solutions, we can use the formula:
Molarity (M) = Moles / Volume (in liters)
For NaH2PO4:
Molarity of NaH2PO4 = (0.200 F x 50.0 mL) / 1000 mL/L = 0.0100 M
For H2PO4- (which is formed by the reaction of HCl with NaH2PO4):
Molarity of H2PO4- = (0.120 F x 50.0 mL) / 1000 mL/L = 0.00600 M
Now we can plug the values into the Henderson-Hasselbalch equation:
pH = pKa + log([A-]/[HA])
= 7.21 + log(0.00600 M / 0.0100 M)
Calculating the ratio [A-]/[HA]:
= 7.21 + log(0.6)
Finally, we can use a calculator to calculate log(0.6), add it to 7.21, and get the pH of the buffer. The final value will depend on the calculation.
Note: Since the Henderson-Hasselbalch equation assumes ideal conditions, this calculation provides an approximation of the buffer pH and may not consider other factors that could affect the pH, such as the ionic strength of the solution and variations in temperature.
millimols H2PO4- = mL x M = approx 10
millimols HCl added = 6
......H2PO4^- + H^+ ==> H3PO4
I......10.......0........0
add............6................
C......-6......-6........6
E......4........0.........6
Substitute the E line into the HH buffer equation and solve for pH.
Post your work if you turn into trouble.