The Foundations of Chemistry

Kb(CN)^1

4

.

.

0

0





1

1

0

0





1

1

4

 0  2.5^10

 5

You should now work Exercises 78 and 80.

EXAMPLE 18-19 Calculations Based on Hydrolysis

Calculate [OH], pH, and the percent hydrolysis for 0.10 Msolutions of (a) sodium acetate,
NaCH 3 COO, and (b) sodium cyanide, NaCN. Both NaCH 3 COO and NaCN are soluble ionic
salts that are completely dissociated in H 2 O. From the text, Kbfor CH 3 COO5.6 10 ^10 ;
from Example 18-18, Kbfor CN2.5 10 ^5.

Plan

We recognize that both NaCH 3 COO and NaCN are salts of strong bases and weak acids. The
anions in such salts hydrolyze to give basic solutions. As we have done before, we first write
the appropriate chemical equation and equilibrium constant expression. Then we complete the
reaction summary, substitute the algebraic representations of equilibrium concentrations into
the equilibrium constant expression, and solve for the unknown concentration(s).

Solution

(a) The overall equation for the reaction of CH 3 COOwith H 2 O and its equilibrium constant
expression are

CH 3 COO
H 2 O 34 CH 3 COOH
OH Kb

5.6 10 ^10

Let xmol/L of CH 3 COOthat hydrolyzes. Then x[CH 3 COOH][OH].

CH 3 COO
H 2 O88nCH 3 COOH
OH

initial 0.10 M 0 M 
 0 M

change due to rxn x M 
x M 
x M

at equil (0.10x) Mx Mx M

Because the value of Kb(5.6 10 ^10 ) is quite small, we know that the reaction does not go
very far. We can assume x0.10, so (0.10x)
0.10; this lets us simplify the equation to

5.6 10 ^10 

(

0

x

.

)

1

(x

0

)

 so x7.5 10 ^6

x 7.5 10 ^6 M[OH] pOH5.12 and pH8.88

The 0.10 MNaCH 3 COO solution is distinctly basic.

% hydrolysis100%100%

 0.0075% hydrolysis

(b) Perform the same kind of calculation for 0.10 MNaCN. Let ymol/L of CNthat
hydrolyzes. Then y[HCN][OH].

7.5 10 ^6 M



0.10 M

[CH 3 COO]hydrolyzed



[CH 3 COO]initial

[CH 3 COOH][OH]



[CH 3 COO]

Kw



Ka

If we did not know Kb, we could use

Ka(HCN)to find Kb(CN).

18-8 Salts of Strong Bases and Weak Acids 779