In: Chemistry
What volume (to the nearest 0.1 mL) of 7.10-M HCl must be added to 0.550 L of 0.350-M K2HPO4 to prepare a pH = 7.50 buffer?
We are working with a Strong Acid/Weak Base solution. The reaction is extensive so we will use up all of the HCl. The reaction table is as follows:
HCl +
HPO42-
----> Cl-
+
H2PO4-
initial:
7.10x 0.1925
0
0
delta:
-7.10x
-7.10x
7.10x 7.10x
equil:
0 0.1925 -
7.10x
7.10x 7.10x
Note: 0.1925 = 0.350 x 0.550
Remember the equation: pH = pKa +
log(nb/na)
The pKa for HPO42- = 7.21
So we get: 7.5 = 7.21 +
log([HPO42-]/[H2PO4-])
7.5 = 7.21 + log((0.1925 - 7.10x)/(7.10x))
Solve for x and this will be the amount of HCl in L to prepare a pH
= 7.60 buffer. Don't forget to convert to mL.
x = 9.19 10-3 L = 9.19 mL