Answer: Mass of monosodium phosphate NaH2PO4 =2,40 g and Mass of disodium phosphate Na2HPO4= 5, 68 g
Explanation:
We identify which one is the Acid and which one is the conjugate base and its molecular weight
In this case: the monosodium phosphate is the Acid [AH] with molecular weight ( 119,98g/mol) and
the disodium phosphate is the conjugate base [ A-] with molecular weight ( 141,96 g/mol)
Review the parameters giving:
+Volume of buffer to prepare : 1L
+pH of buffer to prepare: 7,640
+ Molar Concentration of buffer to prepare: [0,06M] ------- That means [AH] + [ A-] = [0,06M] Equation 1
We use the Henderson-Hasselbach equation to calculate how much acid and base we will need .
pH = pKa + log ([Base]/[Acid]) = pKa + log ([A-]/[AH]) Equation 2
We choose the pKa value that is closest to our pH parameter giving. In our casem we choose pKa 7,198, because is closer to our pH 7,640, and we put them in our Equation 2
pH = pKa + log ([A-]/[AH])
7,640 = 7,198 + log ([A-]/[AH])
0,44= log ([A-]/[AH])
2,77 = [A-]/[AH]
2,77 [AH] = [A-] Equation 3
Now we use Equation 3 in our Equation 1 and make the Sum of equal terms, finally we have the concentration of the Acid
[AH] + [ A-] = [0,06M] Equation 1
2,77 [AH] = [A-] Equation 3
[AH] + 2,77 [AH]= [0,06M]
3,77 [AH] =[0,06M]
[AH] = [0,02M]
And the Concentration of the conjugate base [ A-], can be calculated from Equation 1 :
[AH] + [ A-] = [0,06M]
[0,02M] + [ A-] = [0,06M]
[ A-] = [0,04M]
We calculate the mass of each compound by multiplying its molecular weight with its molar concentration
mass AH = 119,98g/mol X 0,02mol / L = 2,40 g / L
mass A- = 141,96 g/mol X 0,04mol / L = 5, 68 g / L