8
pH AND HYDROLYSIS OF A DOUBLY WEAK SALT KURT EISEMANN Yeshiva University, New York, New York
THE
problem of calculating the pH and hydrolysis of doubly weak salts (i. e., salts of a weak acid and a weak base with different ionization constants) is customarily shunned on account of its comparative complexity. Below an analysis of the order of magnitude of the quantities involved gives a very close approximation and a simplifying rule for solution of specific cases. The Problem. It is required to calculate the pH and percentage hydrolysis of a 0.1 M NH4CN solution.* Notation. C = Concentration of salt originally introduced
-4 = Acid ionization constant = 7 X 10-lo
B
b z
= = = = =
y
=
W
a
=
0.1 M
KHs.HzO and' the cycle of dissociation is repeatedly gone through in diminishing amounts until the relative amounts of ions finally adjust themselves so as to satisfy (11, (21, and (3). Since confusion is likely to arise in visualizing the mutual reactions and arriving a t a clear concept of the quantities takimg part, it is strongly recommended that all such reactions be represented graphically in the following manner, showing the over-all resulting dissociation t,oget,herwith the mutual relat,ions of ion concentrations:
Base ionization constant = 1.8 X 10Water ion product = lo-" F i n d acid anion concentration = [CN-] Final base cation concentration = [NHd+] Final lHfl Final [OH-]
/
Chemical Equilibrittm. The following relations hold between the final ion concentrations at. equilibrium:
These are three equations with four unknowns (x,y, a, b), requiring a fourth relation for solution. In order that the action of hydrolysis may be dealt with in the more familiar concept of ionic dissociation, consider a t first complete bydrolyzation to take place. Since [NH4+] = [CN-1, the amounts of OHand H + used up thereby will be equal. Let also the remaining H + and OH- recombine to form HzO. Xow dissociation takes place, uiz., a moles HCK
--
6 moles NH8.H8O
H'
+C W
NH,+
+ OH-
(4) (5)
in such manner as to satisfy (1) and (2) above. Equation (1) independently would yield [H+] = [CN-] = 40.1 X 7 X 10-'0 = 8.4 X 10-6 moles and (2) independently would yield [OH-] = [NH4+] = moles, so that the 1.8 X 10-5 = 1.3 X combination of (1) and (2) would yield [H+J[OH-I
= 1.1
X 10-8
> lo-''
a
z
w
\
NHlt
y
NHa.H80 C -b
b
The amount of H + and OH- formed by (4) and (5) and recombining to form HnO is w moles. From the diagram it becomes evident that the amount of H + removed from HCN -+ H + CN- in order to form water is equal to the amount of OHremoved from NH3.Hz0 + NH4+ OH- in order to form water, i. e.,
+
+
w = a - z = b - y
m
(6)
This is the fourth relation sought above. It may also he stated as [CN-]
-
[Ht]
=
ISHI+]
-
[OH-]
(7)
Assumption a, b
