How many moles of H2PO4 and HPO4 would be needed to prepare 1.0 L of a 0.01M phosphate buffer with a pH of 6.82?
Use the Henderson-Hasselbalch equation.
pH = pKa + log (base)/(acid).
Thank you, but I guess what my question should have been is what is the acid in this question and what is the base. I would assume that both H2PO4 and HPO4 are acids and phosphate is the base but Im not sure.
To solve this problem, we need to consider the Henderson-Hasselbalch equation, which relates the pH of a buffer solution to the ratio of the concentrations of the acid and its conjugate base:
pH = pKa + log([base]/[acid])
In this case, the acid is H2PO4- (dihydrogen phosphate) and the base is HPO4^2- (monohydrogen phosphate). The pKa of the phosphate buffer system is a value of 7.21 at 25°C.
Given:
pH = 6.82
Volume of buffer solution (V) = 1.0 L
Buffer concentration (C) = 0.01 M
Let's calculate the amount of moles of each species needed:
Step 1: Calculate the ratio of [base] to [acid] using the Henderson-Hasselbalch equation.
pH = pKa + log([base]/[acid])
6.82 = 7.21 + log([HPO4^2-]/[H2PO4-])
Step 2: Solve for the ratio of [HPO4^2-] to [H2PO4-].
log([HPO4^2-]/[H2PO4-]) = 6.82 - 7.21
log([HPO4^2-]/[H2PO4-]) = -0.39
Now we can convert this equation to exponential form:
[HPO4^2-]/[H2PO4-] = 10^(-0.39)
Step 3: Calculate the concentrations of the base ([HPO4^2-]) and the acid ([H2PO4-]).
C = concentration of phosphate buffer solution = 0.01 M
Let's assume the concentration of [HPO4^2-] as x:
[H2PO4-] = C - x
Substituting these values into the equation from Step 2:
x / (C - x) = 10^(-0.39)
Step 4: Solve for x (concentration of [HPO4^2-]).
x = (C * 10^(-0.39)) / (1 + 10^(-0.39))
Step 5: Calculate the moles of each species.
moles of [HPO4^2-] = x * V
moles of [H2PO4-] = (C - x) * V
Substituting the given values:
moles of [HPO4^2-] = [(0.01 M) * 10^(-0.39)] / (1 + 10^(-0.39)) * (1.0 L)
moles of [H2PO4-] = [(0.01 M) * (1 - 10^(-0.39))] / (1 + 10^(-0.39)) * (1.0 L)
Calculating these expressions will give you the moles of HPO4^2- and H2PO4- that would be needed to prepare 1.0 L of a 0.01 M phosphate buffer with a pH of 6.82.
To determine the number of moles of H2PO4 and HPO4 needed in the buffer solution, we first need to understand the Henderson-Hasselbalch equation.
The Henderson-Hasselbalch equation relates the pH of a buffer solution to the ratio of the concentrations of the acidic and basic components of the buffer. In the case of a phosphate buffer, we have two acidic components: H2PO4 (dihydrogen phosphate) and HPO4 (monohydrogen phosphate).
The Henderson-Hasselbalch equation is given by:
pH = pKa + log([A-] / [HA])
Where:
- pH is the desired pH of the buffer solution.
- pKa is the negative logarithm (base 10) of the acid dissociation constant (Ka) of the buffer components.
- [A-] is the concentration of the conjugate base.
- [HA] is the concentration of the acid.
Given that the pH is 6.82, we need to find the pKa values for H2PO4 and HPO4.
The pKa values for H2PO4 and HPO4 are 2.15 and 7.20, respectively.
Using the Henderson-Hasselbalch equation, we can rearrange it to solve for the concentration ratio [A-] / [HA]:
[A-] / [HA] = 10^(pH - pKa)
[A-] / [HA] = 10^(6.82 - 7.20)
[A-] / [HA] = 10^(-0.38)
[A-] / [HA] = 0.459
Since the volume of the buffer solution is 1.0 L and the concentration is 0.01 M, we can calculate the amount of H2PO4 and HPO4 needed:
[H2PO4] + [HPO4] = 0.01 M
Let's assume x represents the concentration of H2PO4 and (0.01 - x) represents the concentration of HPO4.
Using the concentration ratio we found earlier, we can set up an equation:
(0.459) = ([HPO4] / [H2PO4])
(0.459) = (0.01 - x) / x
Cross-multiply:
(0.459)x = 0.01 - x
(1 + 0.459)x = 0.01
1.459x = 0.01
x = 0.01 / 1.459
x ≈ 0.00686 M
Therefore, the concentration of H2PO4 (dihydrogen phosphate) needed is approximately 0.00686 M, and the concentration of HPO4 (monohydrogen phosphate) needed is approximately (0.01 - 0.00686) ≈ 0.00314 M.
To determine the number of moles needed, we multiply the concentrations by the volume (1.0 L):
Number of moles of H2PO4 = 0.00686 mol/L × 1.0 L = 0.00686 mol
Number of moles of HPO4 = 0.00314 mol/L × 1.0 L = 0.00314 mol
Therefore, you would need approximately 0.00686 moles of H2PO4 and 0.00314 moles of HPO4 to prepare 1.0 L of a 0.01 M phosphate buffer with a pH of 6.82.