What volume of 0.170 M K3PO4(aq) must be added to 50.3 mL of 0.103 M K2HPO4(aq) to prepare a buffer solution with a pH of 12.09?
Use the HH equation.
pH = pK3 + log (50.3*0.103)/(0.170*mL)
Solve for mL.
To solve this problem, we need to use the Henderson-Hasselbalch equation for calculating the pH of a buffer solution:
pH = pKa + log([A-]/[HA]), where [A-] is the concentration of the conjugate base and [HA] is the concentration of the weak acid.
First, we need to determine the pKa value for the phosphate buffer system. The pKa value of the phosphate buffer system is the pKa of the acidic component, which is H2PO4-. The pKa value for H2PO4- is 7.21.
Next, we need to calculate the concentrations of the acidic component ([HA]) and the conjugate base ([A-]).
Given:
[HA] = concentration of K2HPO4 = 0.103 M
pH = 12.09
Using the Henderson-Hasselbalch equation, we have:
12.09 = 7.21 + log([A-]/0.103)
Rearranging the equation, we get:
log([A-]/0.103) = 12.09 - 7.21
log([A-]/0.103) = 4.88
Now, let's solve for [A-]:
[A-]/0.103 = 10^4.88
[A-] = 0.103 * 10^4.88
Since K3PO4 is the source of A- ions, we need to find the volume of K3PO4 that should be added to the solution to achieve the desired concentration.
Given:
[K3PO4] = 0.170 M
Let's assume the volume of K3PO4 needed is V.
The moles of [A-] can be calculated as:
moles of [A-] = [A-] * V
Since the moles of [A-] should be equal to the moles of [HA] initially present in 50.3 mL of K2HPO4, we have:
moles of [HA] = [HA] * 50.3 mL (converting mL to L by dividing by 1000)
Now, we can set up the equation:
[A-] * V = [HA] * 50.3 mL
Rearranging the equation to solve for V, we get:
V = [HA] * 50.3 mL / [A-]
Substituting the given values, we get:
V = (0.103 * 50.3 mL) / (0.103 * 10^4.88)
Evaluating the expression, we find:
V ≈ 50.3 mL / 10^4.88
Thus, the volume of 0.170 M K3PO4(aq) that must be added is approximately 0.00005 mL.