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?

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      +     HPO₄^2-    ---->       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 = pK_a + log(n_b/n_a)
The p_Ka for HPO₄^2- = 7.21
So we get: 7.5 = 7.21 + log([HPO₄^2-]/[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

queen_honey_blossom answered 3 years ago

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