Physical Chemistry Third Edition

326 7 Chemical Equilibrium

Instead of using the quadratic formula, we solve by successive approximations within our

successive approximations for the activity coefficients. For the first approximation we neglect

xcompared with 0.100:

x^2 ≈(0.100)(5. 71 × 10 −^10 )

≈ 5. 71 × 10 −^11

x≈

√

5. 71 × 10 −^11  7. 55 × 10 −^6

For the second approximation we replacexby its first approximate value:

x^2 ≈(0. 100 − 2. 39 × 10 −^5 )(5. 71 × 10 −^10 )

≈ 5. 71 × 10 −^11

x≈

√

5. 71 × 10 −^11  7. 55 × 10 −^6

There is no change in the value ofx, so we discontinue successive approximations at this point.

We could have predicted this result from the small size ofxfrom the first approximation. We

now return to the successive approximation scheme for the activity coefficients. The ionic

strength is the same as though the sodium acetate did not react, since every acetate ion that

reacts is replace by a hydroxide ion. The ionic strength isI 0 .100 mol kg−^1. By the Davies

equation:

log 10 (γ±)− 0. 510

⎛

⎜⎝

√

0. 100

1 +

√

0. 100

−(0.30)(0.100)

⎞

⎟⎠

− 0. 1072

γ± 10 −^0.^1072  0. 7813

Assume that this value can be used for both the acetate ion and the hydroxide ion. The two

activity coefficients cancel, so our first value forxcan be used.

m(OH−^1 )/m◦ 7. 55 × 10 −^6

We define the pOH in the same way as the pH:

pOH−log 10 (a(OH−)

−log 10

[

(0.7813)(7. 55 × 10 −^6 )

]

 5. 23

pH+pOHlog 10 (Kw) 14. 00

pH 14. 00 −pOH 8. 77

The conjugate acid of a weak base such as NH 3 hydrolyzes in a way analogous to

the hydrolysis of the acetate ion.

Exercise 7.11

Find the pH of a solution made from 0.100 mol of NH 4 Cl and 1.000 kg of water at 298.15 K