As a technician in a large pharmaceutical research firm, you need to produce 450. mL of 1.00 M a phosphate buffer solution of pH = 7.20. The pKa of H2PO4− is 7.21. You have 2.00 L of 1.00 M KH2PO4 solution and 1.50 L of 1.00 M K2HPO4 solution, as well as a carboy of pure distilled H2O. How much 1.00 M KH2PO4 will you need to make this solution?
Two equations which you solve simultaneously. KH2PO4 is the acid(A); K2HPO4 is the base(B)
pH = pKa + log B/A
So equation 1 is
7.20 = 7.21 + log B/A
Solve for B/A = ?
Equation 2 is
A + B = 1 M. (The problem tells you the buffer is to be 1 M. I like to work in millimols and you want 450 mL of 1 M; therefore,
A + B = 450 mmols.
Solve these two equations
(1) B/A = ?
(2) A + B = 450
for A and B. A is approx 230 mmols and B is approx 220 mmols.
You will need better answers for each since these numbers are estimates (but close estimates)
Now you know M = mmols/mL. You know M for A and mmolsl for A, solve for mL A.
And you know M for B and mmols for B, solve for mL B.
My estimates are aprox 150 mL for B and 230 mL for A. Again these are close estimates. Post your work if you get stuck.
To make a phosphate buffer solution, we need to use the Henderson-Hasselbalch equation:
pH = pKa + log [A-] / [HA]
Where:
pH = desired pH of the buffer solution (7.20)
pKa = dissociation constant of the weak acid (7.21)
[A-] = concentration of the conjugate base (K2HPO4)
[HA] = concentration of the weak acid (KH2PO4)
Step 1: Determine the ratio of [A-] to [HA]
We need to use the Henderson-Hasselbalch equation to find the ratio of [A-] to [HA] that will result in a pH of 7.20.
7.20 = 7.21 + log [A-] / [HA]
Rearranging the equation, we have:
log [A-] / [HA] = 7.20 - 7.21
log [A-] / [HA] = -0.01
Taking the antilog of both sides:
[A-] / [HA] = 10^-0.01
[A-] / [HA] = 0.794
Step 2: Calculate the volume of 2.00 M KH2PO4 needed to achieve the desired concentration of [HA]
Let's assume that x mL of 2.00 M KH2PO4 is needed.
Concentration of KH2PO4 = 2.00 M
Volume of KH2PO4 = x mL
Moles of KH2PO4 = concentration × volume
= (2.00 M) × (x mL) / 1000
= 2x / 1000 moles
Since the stoichiometry of KH2PO4 to KHPO4 is 1:1, the moles of KH2PO4 will be the same as the moles of KHPO4.
Step 3: Calculate the volume of 1.50 M K2HPO4 needed to achieve the desired concentration of [A-]
Let's assume that y mL of 1.50 M K2HPO4 is needed.
Concentration of K2HPO4 = 1.50 M
Volume of K2HPO4 = y mL
Moles of K2HPO4 = concentration × volume
= (1.50 M) × (y mL) / 1000
= 1.5y / 1000 moles
Step 4: Set up an equation using the ratio of [A-] to [HA]
Using the ratio we found earlier:
[A-] / [HA] = 0.794
Plugging in the moles of KHPO4 and K2HPO4:
(1.5y / 1000) / (2x / 1000) = 0.794
1.5y / 2x = 0.794
Simplifying the equation:
y / x = 0.794 × (2 / 1.5)
y / x = 1.059
Step 5: Calculate the required volume of 1.00 M KH2PO4
Now we can set up another equation using the moles of KH2PO4:
Moles of KH2PO4 = concentration × volume
= (1.00 M) × (volume of 1.00 M KH2PO4) / 1000
= (1.00 M) × (450 mL) / 1000
= 0.45 moles
Since the stoichiometry of KH2PO4 to KHPO4 is 1:1, the moles of KH2PO4 will be the same as the moles of KHPO4:
0.45 moles = 2x / 1000
x = 0.45 × 1000 / 2
x = 225 mL
Therefore, you will need 225 mL of the 2.00 M KH2PO4 solution to make the 450 mL of 1.00 M phosphate buffer solution with a pH of 7.20.
To solve this problem, we need to determine the amounts of KH2PO4 and K2HPO4 needed to prepare a 450 mL solution of a phosphate buffer with a pH of 7.20.
First, let's understand the principle behind making a phosphate buffer. Phosphate buffers consist of a mixture of two components: a weak acid (H2PO4-) and its conjugate base (HPO4^2-). By adjusting the respective concentrations of these two components, we can control the pH of the buffer solution. The pKa of the weak acid is an important value in determining the pH of the buffer.
In this case, the pKa of H2PO4- is given as 7.21, which is very close to the desired pH of 7.20. This means that at pH 7.20, the concentrations of the weak acid (H2PO4-) and its conjugate base (HPO4^2-) will be almost equal.
To calculate the quantities needed, we can use the Henderson-Hasselbalch equation:
pH = pKa + log([A-]/[HA])
Where pH is the desired pH, pKa is the acid dissociation constant, [A-] is the concentration of the conjugate base, and [HA] is the concentration of the weak acid.
Let's denote the concentration of KH2PO4 we need to calculate as x. We know the following values:
[H2PO4-] = 1.00 M (from the 2.00 L of 1.00 M KH2PO4 solution)
[HPO4^2-] = 1.00 M (from the 1.50 L of 1.00 M K2HPO4 solution)
pKa = 7.21
pH = 7.20
Using the Henderson-Hasselbalch equation, we can set up the following equation:
7.20 = 7.21 + log([HPO4^2-]/x)
To simplify the calculation, we can take the antilog of both sides:
10^(7.20 - 7.21) = [HPO4^2-]/x
10^(-0.01) = [HPO4^2-]/x
0.794328 = [HPO4^2-]/x
Now we can solve for [HPO4^2-]:
[HPO4^2-] = 0.794328 * x
Since [HPO4^2-] equals 1.00 M (from the 1.50 L of 1.00 M K2HPO4 solution), we can substitute it into the equation:
1.00 M = 0.794328 * x
Now we can solve for x, which represents the amount (in L) of KH2PO4 solution needed:
x = 1.00 M / 0.794328
x ≈ 1.26 L
Therefore, you will need approximately 1.26 L of the 1.00 M KH2PO4 solution to make the 450 mL phosphate buffer solution with a pH of 7.20.