Question

In: Chemistry

What volume (to the nearest 0.1 mL) of 7.10-M HCl must be added to 0.550 L...

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?

Solutions

Expert Solution

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


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