Phosphoric acid is a triprotic acid with the following pKa values:
pKa1=2.148 pKa2=7.198 pKa3=12.375
You wish to prepare 1.000 L of a 0.0200 M phosphate buffer at pH 7.11. To do this, you choose to use mix the two salt forms involved in the second ionization, NaH2PO4 and Na2HPO4, in a 1.000 L volumetric flask and add water to the mark. What mass of each salt will you add to the mixture?
To determine how much of each salt to add, we need to consider the Henderson-Hasselbalch equation, which relates the ratio of the salt forms and their pKa values to the desired pH of the buffer solution:
pH = pKa + log [salt form 2]/[salt form 1]
In this case, we want the pH of the buffer to be 7.11. The two salt forms involved in the second ionization are NaH2PO4 (salt form 1) and Na2HPO4 (salt form 2). The pKa value for the second ionization is 7.198.
We can rearrange the Henderson-Hasselbalch equation to solve for the ratio [salt form 2]/[salt form 1]:
[salt form 2]/[salt form 1] = 10^(pH - pKa)
Plugging in the values, we get:
[salt form 2]/[salt form 1] = 10^(7.11 - 7.198)
= 10^(-0.088)
≈ 0.8697
This tells us that the ratio of the two salt forms should be approximately 0.8697:1.
Next, we need to calculate the moles of each salt required to prepare the buffer solution. We know the volume (1.000 L) and molarity (0.0200 M) of the buffer solution, so we can use the formula:
moles = volume × molarity
For NaH2PO4:
moles NaH2PO4 = 1.000 L × 0.0200 M = 0.0200 mol
Since the ratio of [salt form 2]/[salt form 1] is approximately 0.8697:1, we can calculate the moles of Na2HPO4 based on the moles of NaH2PO4:
moles Na2HPO4 = 0.8697 × 0.0200 mol ≈ 0.0174 mol
Now, we can calculate the mass of each salt required using their respective molar masses.
The molar mass of NaH2PO4 is:
(mass of Na + mass of H2PO4) = (22.99 g/mol + 1.008 g/mol × 2 + 31.00 g/mol + 16.00 g/mol × 4)
= 120.00 g/mol
mass NaH2PO4 = moles NaH2PO4 × molar mass NaH2PO4
= 0.0200 mol × 120.00 g/mol
= 2.40 g
The molar mass of Na2HPO4 is:
(mass of Na × 2 + mass of HPO4) = (22.99 g/mol × 2 + 31.00 g/mol + 16.00 g/mol × 4)
= 142.00 g/mol
mass Na2HPO4 = moles Na2HPO4 × molar mass Na2HPO4
= 0.0174 mol × 142.00 g/mol
≈ 2.47 g
Therefore, you will need to add approximately 2.40 g of NaH2PO4 and 2.47 g of Na2HPO4 to the mixture.