CK-12-Chemistry Intermediate

21.5. Salt Solutions http://www.ck12.org

Kb= 1. 4 × 10 −^11 =

(x)(x)

0. 953 −x

=

x^2

0. 953 −x

≈

x^2

0. 953

x= [OH−] =

√

1. 4 × 10 −^11 ( 0. 953 ) = 3. 65 × 10 −^6 M

pOH = -log(3.65× 10 −^6 ) = 5.44

pH = 14 - 5.44 = 8.56

Step 3: Think about your result.

The solution is slightly basic due to the hydrolysis of the fluoride ion.

Practice Problem

1. The Kbof the acetate ion (CH 3 COO−) is 5.6× 10 −^10. Calculate the pH of a 0.200 M solution of sodium
   acetate.

Salts that are derived from the neutralization of a weak acid (HF) by a strong base (NaOH) will always produce salt
solutions that are basic. Now we will see what happens with a salt that is formed from the reaction of a strong acid
with a weak base.

Salts That Form Acidic Solutions

Ammonium chloride (NH 4 Cl) is a salt that is formed when the strong acid HCl is neutralized by the weak base NH 3.
Ammonium chloride is soluble in water. The chloride ion cannot undergo an acid-base reaction with water because
it is the conjugate base of the strong acid HCl. In other words, the Cl−ion cannot accept a proton from water to form
HCl and OH−, as the fluoride ion did in the previous section. However, the ammonium ion is capable of reacting
slightly with water by acting as a proton donor.

NH+ 4 (aq)+H 2 O(l)⇀↽H 3 O+(aq)+NH 3 (aq)

The production of hydronium ions causes the resulting solution to be acidic. The pH of a solution of ammonium
chloride can be found in a very similar way to the pH of the sodium fluoride solution in Sample Problem 21.7.
However, since ammonium chloride is acting as an acid, it is necessary to know the Kaof NH 4 +, which is 5.6
× 10 −^10. As an example, we will find the pH of a 2.00 M solution of NH 4 Cl. Because the NH 4 Cl completely
dissociates in water, the concentration of the ammonium ion is also 2.00 M.

NH 4 Cl(s)→NH+ 4 (aq)+Cl−(aq)

Again, an ICE table is set up in order to solve for the concentration of the hydronium (or H+) ion produced (Table
21.11).

TABLE21.11:ICE Table

Concentrations [NH 4 +] [H+] [NH 3 ]

Initial 2.00 0 0

Change −x +x +x

Equilibrium 2.00−x x x