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?
You need two equations and solve them simultaneously.
1000 mL x 0.02M = 20 millimols.
mols acid + mols base = 20
pH = pKa + log(base)/(acid)
7.0 = pK2 + log (base)/(acid) and
you solve for (base)/acd) = ?
Then solve the two equations for mols acid and mols base, convert mols each to grams for 1 L. Post your work if you get stuck.
To determine the mass of each salt that needs to be added to the mixture, we need to consider the Henderson-Hasselbalch equation for a buffer solution:
pH = pKa + log([A-]/[HA])
Where pH is the desired pH of the buffer, pKa is the dissociation constant of the acid, [A-] is the concentration of the conjugate base (Na2HPO4 in this case), and [HA] is the concentration of the acid (NaH2PO4 in this case).
Rearranging the equation, we get:
[A-]/[HA] = 10^(pH - pKa)
Now, let's calculate the ratio of [A-]/[HA] for the given pH:
[A-]/[HA] = 10^(7.11 - 7.198) = 10^(-0.088) = 0.756
Since we want to prepare a 0.0200 M phosphate buffer, we can choose any concentration for [HA], let's assume it to be x M. Therefore, [A-] would be 0.756x M.
Now, the total concentration of the two salts combined in the buffer solution should be 0.0200 M. Therefore:
x + 0.756x = 0.0200
1.756x = 0.0200
x = 0.0200 / 1.756
x = 0.0114 M
Now we know that the concentration of NaH2PO4 ([HA]) should be 0.0114 M, and the concentration of Na2HPO4 ([A-]) should be 0.756 * 0.0114 M = 0.00864 M.
To calculate the mass of each salt, we need to use their respective molar masses:
NaH2PO4 Molar Mass = 22.99 g/mol (Na) + 1.01 g/mol (H) + 2 * 1.01 g/mol (O) + 31.00 g/mol (P) = 119.98 g/mol
Na2HPO4 Molar Mass = 2 * 22.99 g/mol (Na) + 1.01 g/mol (H) + 4 * 1.01 g/mol (O) + 31.00 g/mol (P) = 141.96 g/mol
Finally, we can calculate the mass of each salt using their respective molar masses and concentrations:
Mass of NaH2PO4 = 0.0114 M * 1.000 L * 119.98 g/mol = 1.367 g
Mass of Na2HPO4 = 0.00864 M * 1.000 L * 141.96 g/mol = 1.229 g
Therefore, you will need to add approximately 1.367 g of NaH2PO4 and 1.229 g of Na2HPO4 to prepare the 1.000 L of the 0.0200 M phosphate buffer at pH 7.11.