Finding the
pH of a Weak
Base
Here we’ll go through the steps necessary to find the pH of a weak base with a
given concentration.
Find the pH of a 0.35 M solution
of NaNO2.
We’re asked to find the pH of a 0.35 M solution of the salt sodium nitrite, NaNO2.
Find the pH of a 0.35 M solution
of NaNO2.
1. Dissociate the salt into it’s ions and discard any
spectator ions.
A general approach to finding the pH of a salt solution is outlined here. We start by
dissociating the salt into its individual ions and discarding any spectator ions.
Find the pH of a 0.35 M solution
of NaNO2.
1. Dissociate the salt into it’s ions and discard any
spectator ions.
2. Identify remaining ions as acids or bases and as
strong or weak.
Next, we look at the remaining ion or ions and identify them as acids or bases and
as strong or weak. We use location on the acid table for this.
Find the pH of a 0.35 M solution
of NaNO2.
1. Dissociate the salt into it’s ions and discard any
spectator ions.
2. Identify remaining ions as acids or bases and as
strong or weak.
3. If it is weak, write the hydrolysis equilibrium
equation.
If the ion acts as a WEAK acid or base we write its equilibrium equation for
hydrolysis, the ion’s reaction with water.
Find the pH of a 0.35 M solution
of NaNO2.
1. Dissociate the salt into it’s ions and discard any
spectator ions.
2. Identify remaining ions as acids or bases and as
strong or weak.
3. If it is weak, write the hydrolysis equilibrium
equation.
4. Using an ICE table, calculate [H3O+] or [OH–].
If it’s a weak acid, we use an ICE table and its Ka expression to calculate the
hydronium ion concentration,
Find the pH of a 0.35 M solution
of NaNO2.
1. Dissociate the salt into it’s ions and discard any
spectator ions.
2. Identify remaining ions as acids or bases and as
strong or weak.
3. If it is weak, write the hydrolysis equilibrium
equation.
4. Using an ICE table, calculate [H3O+] or [OH–].
and if it’s a weak base, we use an ICE table and its Kb expression to calculate the
hydroxide ion concentration.
Find the pH of a 0.35 M solution
of NaNO2.
1. Dissociate the salt into it’s ions and discard any
spectator ions.
2. Identify remaining ions as acids or bases and as
strong or weak.
3. If it is weak, write the hydrolysis equilibrium
equation.
4. Using an ICE table, calculate [H3O+] or [OH–].
5. Find the pH.
Finally we convert hydronium or hydroxide concentration to pH.
Find the pH of a 0.35 M solution
of NaNO2.
Because NaNO2 is a salt, we’ll start by dissociating it into its individual ions.
Find the pH of a 0.35 M solution
of NaNO2.


NaNO 2(aq)  Na(aq)
 NO 2(aq)
Neutral
Spectator
Which are Na+ and NO2 minus ions.
Weak
Base
Find the pH of a 0.35 M solution
of NaNO2.


NaNO 2(aq)  Na(aq

NO
)
2(aq)
Neutral
Spectator
Na+ is an alkali metal cation, so it is a neutral spectator.
Weak
Base
Find the pH of a 0.35 M solution
of NaNO2.


NaNO 2(aq)  Na(aq

NO
)
2(aq)
Neutral
Spectator
And we can discard it.
Weak
Base
Find the pH of a 0.35 M solution
of NaNO2.

NaNO 2(aq)  Na(aq)
 NO2(aq)
?
In order to determine what NO2 minus acts as, we look for it on the acid table.
Weak
Base
It’s at this location on the right side of the table, so it’s a weak base.


NaNO 2(aq)  Na(aq)
 NO2(aq)
Weak
Base
Find the pH of a 0.35 M solution of NaNO2.
Because NO2 minus is a weak base
K b(NO ) 
2

Kw
K a(conjugate acid)
Kw

NaNO 2(aq)  Na(aq)
 NO2(aq)
K a(HNO2 )
1.00  10 14
11


2.17

10
4
4.6  10
Find the pH of a 0.35 M solution of NaNO2.
We will need to find the value of its Kb
Weak
Base
K b(NO ) 
2

Kw
K a(conjugate acid)
Kw
K a(HNO2 )
1.00  10 14
11


2.17

10
4
4.6  10
Find the pH of a 0.35 M solution of NaNO2.
Remember the formula we can use for Kb of NO2 minus is Kb = Kw over the Ka of
its conjugate acid.
The conjugate
acid of NO2– is
HNO2
Looking at the table, we see that the conjugate acid of NO2 minus is HNO2.
K b(NO ) 
2
K b(NO ) 
2
Kw
K a(conjugate acid)
Kw
K a(HNO2 )
1.00  10 14
11


2.17

10
4
4.6  10
Find the pH of a 0.35 M solution of NaNO2.
So the Kb of NO2 minus is Kw over the Ka of HNO2.
Ka of
HNO2
The Ka for HNO2 is shown on the table here, and its 4.6 × 10-4.
K b(NO ) 
2
K b(NO ) 
2
Kw
K a(conjugate acid)
Kw
K a(HNO2 )
1.00  10 14
11


2.17

1
0
4
4.6  10
Find the pH of a 0.35 M solution of NaNO2.
Now we can substitute. We know that Kw is 1.00 × 10-14,
K b(NO ) 
2
K b(NO ) 
2
Kw
K a(conjugate acid)
Kw
K a(HNO2 )
1.00  10 14
11


2.17

10
4
4.6  10
Find the pH of a 0.35 M solution of NaNO2.
And the Ka of HNO2 is 4.6 × 10-4.
K b(NO ) 
2

K b(NO )
2
Kw
K a(conjugate acid)
Kw
K a(HNO2 )
1.00  10 14
11


2.
17

10
4
4.6  10
Find the pH of a 0.35 M solution of NaNO2.
1 × 10-14 divided by 4.6 × 10-4 comes out to 2.17 × 10-11. Even though the Ka we used has only 2 significant
figures, we’ll express the Kb to 3 significant figures and round to 2 significant figures in the final answer.
K b(NO ) 
2

K b(NO )
2
Kw
Kb(NO )  2.17  1011
2
K a(conjugate acid)
Kw
K a(HNO2 )
1.00  10 14
11


2.
17

10
4
4.6  10
Find the pH of a 0.35 M solution of NaNO2.
So we’ll make a note up here that the Kb of NO2 minus is 2.17 times 10-11.
K b(NO )  2.17  10 11
2
NO 2(aq)  H 2O( l )

HNO2(aq)  OH(aq)
Weak
Base
Find the pH of a 0.35 M solution of NaNO2.
Because NO2 minus is a weak base, we can write its hydrolysis equilibrium
equation.
K b(NO )  2.17  10 11
2
NO 2(aq)  H 2O( l )

HNO 2(aq)  OH(aq)
Find the pH of a 0.35 M solution of NaNO2.
Hydrolysis means we add it to water.
H+
NO 2(aq)  H 2O( l )
K b(NO )  2.17  10 11
2

HNO 2(aq)  OH(aq)
Weak
Base
Find the pH of a 0.35 M solution of NaNO2.
Because it’s a base, it will accept a proton from water.
K b(NO )  2.17  10 11
H+
NO 2(aq)  H 2O( l )
2

HNO 2(aq)  OH(aq)
Find the pH of a 0.35 M solution of NaNO2.
Forming HNO2
K b(NO )  2.17  10 11
H+
NO 2(aq)  H 2O( l )
2

HNO 2(aq)  OH(a
q)
Find the pH of a 0.35 M solution of NaNO2.
And OH minus.
K b(NO )  2.17  10 11
2
NO 2(aq)  H 2O( l )

HNO 2(aq)  OH(a
q)
?
Find the pH of a 0.35 M solution of NaNO2.
On the way to finding pH, we can start by finding the hydroxide ion concentration.
Because NO2 minus is a WEAK base, we do this using an ICE table.
K b(NO )  2.17  10 11
2
NO 2(aq)  H 2O( l )

HNO2(aq)  OH(aq)
[I]
[C]
[E]
Find the pH of a 0.35 M solution of NaNO2.
So we draw an ICE table underneath this equilibrium equation.
K b(NO )  2.17  10 11
2
NO 2(aq)  H 2O( l )

HNO2(aq)  OH(aq)
[I]
[C]
[E]
Find the pH of a 0.35 M solution of NaNO2.
Water is a liquid so we don’t include it in our calculations. We’ll colour the column
below water blue.
K b(NO )  2.17  10 11
2
NO 2(aq)  H 2O( l )
[I] 0.35
[C] –x
[E] 0.35–x

HNO2(aq)  OH(aq)
0
+x
x
Find the pH of a 0.35 M solution of NaNO2.
Now we’ll fill in what we can in the initial concentration row.
0
+x
x
K b(NO )  2.17  10 11
2
NO 2(aq)  H 2O( l )
[I] 0.35
[C] –x
[E] 0.35–x

HNO2(aq)  OH(aq)
0
+x
x
0
+x
x
Find the pH of a 0.35 M solution of NaNO2.
The initial concentration of NO2 minus is the same as the initial concentration of
NaNO2, and it’s equal to 0.35 molar, so we’ll write 0.35 in here.
K b(NO )  2.17  10 11
2
NO 2(aq)  H 2O( l )
[I] 0.35
[C] –x
[E] 0.35–x

HNO2(aq)  OH(aq)
0
+x
x
0
+x
x
Find the pH of a 0.35 M solution of NaNO2.
Before hydrolysis occurs, we can say that the concentrations of HNO2 and OH
minus are both zero.
K b(NO )  2.17  10 11
2
NO 2(aq)  H 2O( l )

HNO2(aq)  OH(aq)
[I] 0.35
[C] –x
[E] 0.35–x
0
+x
x
Find the pH of a 0.35 M solution of NaNO2.
Now, we’ll fill in the change in concentration row.
0
+x
x
K b(NO )  2.17  10 11
2
NO 2(aq)  H 2O( l )
[I] 0.35
[C] –x
[E] 0.35–x

HNO2(aq)  OH(aq)
0
+x
x
0
+x
x
Find the pH of a 0.35 M solution of NaNO2.
Because the concentrations of the products are zero, in order to compensate…
K b(NO )  2.17  10 11
2
Moves
to
NO 2(aq)  H 2O( l )
[I] 0.35
[C] –x
[E] 0.35–x
the
Right

HNO2(aq)  OH(aq)
0
+x
x
Find the pH of a 0.35 M solution of NaNO2.
The reaction will move to the right.
0
+x
x
K b(NO )  2.17  10 11
2
Moves
to
NO 2(aq)  H 2O( l )
[I] 0.35
[C] –x
[E] 0.35–x
the
Right

HNO2(aq)  OH(aq)
0
+x
x
Find the pH of a 0.35 M solution of NaNO2.
So the concentrations of HNO2 and OH minus will both increase
0
+x
x
K b(NO )  2.17  10 11
2
Moves
to
NO 2(aq)  H 2O( l )
[I] 0.35
[C] –x
[E] 0.35–x
the
Right

HNO2(aq)  OH(aq)
0
+x
x
0
+x
x
Find the pH of a 0.35 M solution of NaNO2.
We’re not given any values at equilibrium and these both have a coefficient of 1 in
the equation, we’ll state that these both increase by x
K b(NO )  2.17  10 11
2
Moves
to
NO 2(aq)  H 2O( l )
[I] 0.35
[C] –x
[E] 0.35–x
the
Right

HNO2(aq)  OH(aq)
0
+x
x
0
+x
x
Find the pH of a 0.35 M solution of NaNO2.
Because the reaction is moving to the right, the concentration of the reactant, NO2
minus, will decrease, so we’ll write a minus sign here.
K b(NO )  2.17  10 11
2
Moves
to
NO 2(aq)  H 2O( l )
[I] 0.35
[C] –x
[E] 0.35–x
the
Right

HNO2(aq)  OH(aq)
0
+x
x
Find the pH of a 0.35 M solution of NaNO2.
Like the HNO2 and the OH minus, the coefficient on NO2 minus is 1,
0
+x
x
K b(NO )  2.17  10 11
2
Moves
to
NO 2(aq)  H 2O( l )
[I] 0.35
[C] –x
[E] 0.35–x
the
Right

HNO2(aq)  OH(aq)
0
+x
x
Find the pH of a 0.35 M solution of NaNO2.
So we can say that the concentration of NO2 minus goes down by x.
0
+x
x
K b(NO )  2.17  10 11
2
NO 2(aq)  H 2O( l )
[I] 0.35
[C] –x
[E] 0.35–x

HNO2(aq)  OH(aq)
0
+x
x
Find the pH of a 0.35 M solution of NaNO2.
Now we can fill in the equilibrium concentration row.
0
+x
x
K b(NO )  2.17  10 11
2
NO 2(aq)  H 2O( l )
[I] 0.35
[C] –x
[E] 0.35–x

HNO2(aq)  OH(aq)
0
+x
x
Find the pH of a 0.35 M solution of NaNO2.
We’ll start with the OH minus. It will be 0 plus x
0
+x
x
K b(NO )  2.17  10 11
2
NO 2(aq)  H 2O( l )
[I] 0.35
[C] –x
[E] 0.35–x

HNO2(aq)  OH(aq)
0
+x
x
Find the pH of a 0.35 M solution of NaNO2.
Which is equal to x
0
+x
x
K b(NO )  2.17  10 11
2
NO 2(aq)  H 2O( l )
[I] 0.35
[C] –x
[E] 0.35–x

HNO2(aq)  OH(aq)
0
+x
x
Find the pH of a 0.35 M solution of NaNO2.
Similarly, the equilibrium concentration of HNO2 will be 0 + x
0
+x
x
K b(NO )  2.17  10 11
2
NO 2(aq)  H 2O( l )
[I] 0.35
[C] –x
[E] 0.35–x

HNO2(aq)  OH(aq)
0
+x
x
Find the pH of a 0.35 M solution of NaNO2.
Which is equal to x
0
+x
x
K b(NO )  2.17  10 11
2
NO 2(aq)  H 2O( l )
[I] 0.35
[C] –x
[E] 0.35–x

HNO2(aq)  OH(aq)
0
+x
x
Find the pH of a 0.35 M solution of NaNO2.
Now, we’ll look at the NO2 minus. It started out as 0.35 molar
0
+x
x
K b(NO )  2.17  10 11
2
NO 2(aq)  H 2O( l )
[I] 0.35
[C] –x
[E] 0.35–x

HNO2(aq)  OH(aq)
0
+x
x
Find the pH of a 0.35 M solution of NaNO2.
And it went down by x,
0
+x
x
K b(NO )  2.17  10 11
2
NO 2(aq)  H 2O( l )
[I] 0.35
[C] –x
[E] 0.35–x

HNO2(aq)  OH(aq)
0
+x
x
Find the pH of a 0.35 M solution of NaNO2.
So its equilibrium concentration is 035 minus x
0
+x
x
K b(NO )  2.17  10 11
2
NO 2(aq)  H 2O( l )
[I] 0.35
[C] –x
[E] 0.35–x

HNO2(aq)  OH(aq)
0
+x
x
Find the pH of a 0.35 M solution of NaNO2.
So now we have the equilibrium concentrations of all species.
0
+x
x
K b(NO )  2.17  10 11
2
NO 2(aq)  H 2O( l )
[I] 0.35
[C] –x
[E] 0.35–x

HNO2(aq)  OH(aq)
0
+x
x
Find the pH of a 0.35 M solution of NaNO2.
Notice that the equilibrium concentration of OH minus is x.
0
+x
x
 OH    x K b(NO )  2.17  10 11


2
NO 2(aq)  H 2O( l )
[I] 0.35
[C] –x
[E] 0.35–x

HNO2(aq)  OH(aq)
0
+x
x
Find the pH of a 0.35 M solution of NaNO2.
We’ll make a note of that up here.
0
+x
x
OH   x K b(NO )  2.17  10 11


2
NO 2(aq)  H 2O( l )

HNO2(aq)  OH(aq)
[I] 0.35
[C] –x
[E] 0.35–x
0
+x
x
Find the pH of a 0.35 M solution of NaNO2.
At this point, we need to solve for the value of x.
0
+x
x


 HNO2  OH 
Kb 
 NO 2 


OH   x K b(NO )  2.17  10 11


2
x2
Kb 
 0.35  x 
x2
Kb 
0.35
Find the pH of a 0.35 M solution of NaNO2.
We start by using the hydrolysis equilibrium equation to write the Kb expression for NO2 minus, which is
the concentration of HNO2 times the concentration of OH minus over the concentration of NO2 minus.


 HNO2  OH 
Kb 
 NO 2 


OH   x K b(NO )  2.17  10 11


2
x2
Kb 
Weak
 x
 0.35Base
x2
Kb 
0 .3 5
Find the pH of a 0.35 M solution of NaNO2.
Remember, because we’re dealing with the hydrolysis of a weak BASE, the
expression is called Kb rather than Ka.


 HNO2  OH 
Kb 
 NO 2 


OH   x K b(NO )  2.17  10 11


2
x2
Kb 
 0.35  x 
x2
Kb 
0.35
Find the pH of a 0.35 M solution of NaNO2.
Now we’ll insert equilibrium concentrations into the Kb expression in order to
solve for x.


 HNO2  OH 
Kb 
 NO 2 


OH   x K b(NO )  2.17  10 11


2
x2
Kb 
 0.35  x 
x2
Kb 
0.35
Find the pH of a 0.35 M solution of NaNO2.
The concentration of HNO2 and OH minus are both equal to x,


 HNO2  OH 
Kb 
 NO 2 


OH   x K b(NO )  2.17  10 11


2
x2
Kb 
 0.35  x 
x2
Kb 
0.35
Find the pH of a 0.35 M solution of NaNO2.
so their product in the Kb expression is x times x, or x squared


 HNO2  OH 
Kb 
 NO 2 


OH   x K b(NO )  2.17  10 11


2
x2
Kb 
 0.35  x 
x2
Kb 
0.35
Find the pH of a 0.35 M solution of NaNO2.
And the equilibrium concentration of NO2 minus is 0.35 minus x,


 HNO2  OH 
Kb 
 NO 2 


OH   x K b(NO )  2.17  10 11


2
x2
Kb 
 0.35  x 
x2
Kb 
0.35
Find the pH of a 0.35 M solution of NaNO2.
so we’ll substitute that in here for the concentration of NO2 minus.


 HNO2  OH 
Kb 
 NO 2 


OH   x K b(NO )  2.17  10 11


2
x2
Kb 
 0.35  x 
x2
Kb 
0.35
Assume
0.35–x  0.35
Find the pH of a 0.35 M solution of NaNO2.
In order to avoid a quadratic equation, we assume that 0.35 minus x is almost equal
to 0.35.


 HNO2  OH 
Kb 
 NO 2 


OH   x K b(NO )  2.17  10 11


2
x2
Kb 
 0.35  x 
x2
Kb 
0.35
Assume
0.35–x  0.35
Find the pH of a 0.35 M solution of NaNO2.
We know this assumption is valid because the Kb of NO2 minus is very small. That means the
amount it deceases by will be very small compared to its initial concentration


 HNO2  OH 
Kb 
 NO 2 


OH   x K b(NO )  2.17  10 11


2
x2
Kb 
 0.35  x 
x2
Kb 
0.35
Assume
0.35–x  0.35
Find the pH of a 0.35 M solution of NaNO2.
Using this assumption, we take the x out of the denominator


 HNO2  OH 
Kb 
 NO 2 


OH   x K b(NO )  2.17  10 11


2
x2
Kb 
 0.35  x 
x2
Kb 
0.35
Find the pH of a 0.35 M solution of NaNO2.
And we get that Kb is approximately equal to x squared divided by 0.35.


 HNO2  OH 
Kb 
 NO 2 


x2
Kb 
 0.35  x 
x2
Kb 
0.35
OH   x K b(NO )  2.17  10 11


2
x 2  0.35 K b
 OH    x  0.35 K b



 OH    0.35 2.17  10 11



 OH    7.595  10 12  2.76  10 6 M




pOH   log 2.76  10 6  5.559
pH  14.000  5.559  8.44
Find the pH of a 0.35 M solution of NaNO2.
Rearranging this equation gives us x squared = 0.35 times Kb


 HNO2  OH 
Kb 
 NO 2 


x2
Kb 
 0.35  x 
x2
Kb 
0.35
OH   x K b(NO )  2.17  10 11


2
x 2  0.35 K b
Take
square
root of
both sides
 OH    x  0.35 K b



 OH    0.35 2.17  10 11



 OH    7.595  10 12  2.76  10 6 M




pOH   log 2.76  10 6  5.559
pH  14.000  5.559  8.44
Find the pH of a 0.35 M solution of NaNO2.
Taking the square root of both sides gives us x = the square root of 0.35 times kb


 HNO2  OH 
Kb 
 NO 2 


x2
Kb 
 0.35  x 
x2
Kb 
0.35
 OH    x K b(NO )  2.17  10 11


2
x 2  0.35 K b
 OH    x  0.35 K b



 OH    0.35 2.17  10 11



 OH    7.595  10 12  2.76  10 6 M




pOH   log 2.76  10 6  5.559
pH  14.000  5.559  8.44
Find the pH of a 0.35 M solution of NaNO2.
Remember from our ice table, that x is equal to the hydroxide ion concentration at
equilibrium.


 HNO2  OH 
Kb 
 NO 2 


x2
Kb 
 0.35  x 
x2
Kb 
0.35
 OH    x K b(NO )  2.17  10 11


2
x 2  0.35 K b
 OH    x  0.35 K b



 OH    0.35 2.17  10 11



 OH    7.595  10 12  2.76  10 6 M




pOH   log 2.76  10 6  5.559
pH  14.000  5.559  8.44
Find the pH of a 0.35 M solution of NaNO2.
So we can say that the hydroxide ion concentration is equal to x, which is equal to
the square root of 0.35 kb


 HNO2  OH 
Kb 
 NO 2 


x2
Kb 
 0.35  x 
x2
Kb 
0.35
 OH    x K b(NO )  2.17  10 11


2
x 2  0.35 K b
 OH    x  0.35 K b



 OH    0.35 2.17  10 11



 OH    7.595  10 12  2.76  10 6 M




pOH   log 2.76  10 6  5.559
pH  14.000  5.559  8.44
Find the pH of a 0.35 M solution of NaNO2.
Or more simply, [OH-] is equal to the square root of 0.35 kb


 HNO2  OH 
Kb 
 NO 2 


x2
Kb 
 0.35  x 
x2
Kb 
0.35
K b(NO )  2.17  10 11
2
x 2  0.35 K b
 OH    x  0.35 K b



 OH    0.35 2.17  10 11



 OH    7.595  10 12  2.76  10 6 M




pOH   log 2.76  10 6  5.559
pH  14.000  5.559  8.44
Find the pH of a 0.35 M solution of NaNO2.
Now we’ll substitute 2.17 × 10-11 in for the value of Kb in the equation


 HNO2  OH 
Kb 
 NO 2 


x2
Kb 
 0.35  x 
x2
Kb 
0.35
K b(NO )  2.17  10 11
2
x 2  0.35 K b
 OH    x  0.35 K b



 OH    0.35 2.17  10 11



 OH    7.595  10 12  2.76  10 6 M




pOH   log 2.76  10 6  5.559
pH  14.000  5.559  8.44
Find the pH of a 0.35 M solution of NaNO2.
0.35 times 2.17 × 10-11 is equal to 7.595 × 10-12.


 HNO2  OH 
Kb 
 NO 2 


x2
Kb 
 0.35  x 
x2
Kb 
0.35
x 2  0.35 K b
 OH    x  0.35 K b



 OH    0.35 2.17  10 11



 OH    7.595  10 12  2.76  10 6 M




pOH   log 2.76  10 6  5.559
pH  14.000  5.559  8.44
Find the pH of a 0.35 M solution of NaNO2.
Taking the square root of 7.595 × 10-12 gives us 2.76 × 10-6.


 HNO2  OH 
Kb 
 NO 2 


x2
Kb 
 0.35  x 
x2
Kb 
0.35
x 2  0.35 K b
 OH    x  0.35 K b



 OH    0.35 2.17  10 11



 OH    7.595  10 12  2.76  10 6 M




pOH   log 2.76  10 6  5.559
pH  14.000  5.559  8.44
Find the pH of a 0.35 M solution of NaNO2.
Because we now have a value for the concentration of hydroxide, we add the unit M
for molarity.


 HNO2  OH 
Kb 
 NO 2 


x2
Kb 
 0.35  x 
x2
Kb 
0.35
x 2  0.35 K b
 OH    x  0.35 K b



 OH    0.35 2.17  10 11



 OH    7.595  10 12  2.76  10 6 M




pOH   log 2.76  10 6  5.559
pH  14.000  5.559  8.44
Find the pH of a 0.35 M solution of NaNO2.
pOH is the negative log of the hydroxide ion concentration, which is the negative
log of 2.76 × 10-6.


 HNO2  OH 
Kb 
 NO 2 


x2
Kb 
 0.35  x 
x2
Kb 
0.35
x 2  0.35 K b
 OH    x  0.35 K b



 OH    0.35 2.17  10 11



 OH    7.595  10 12  2.76  10 6 M




pOH   log 2.76  10 6  5.559
pH  14.000  5.559  8.44
Find the pH of a 0.35 M solution of NaNO2.
And that comes out to 5.559. We’ll stick with 3 significant figures here and round
off to 2 in the last step.


 HNO2  OH 
Kb 
 NO 2 


x2
Kb 
 0.35  x 
x2
Kb 
0.35
x 2  0.35 K b
 OH    x  0.35 K b



 OH    0.35 2.17  10 11



 OH    7.595  10 12  2.76  10 6 M




pOH   log 2.76  10 6  5.559
pH  14.000  5.559  8.44
Find the pH of a 0.35 M solution of NaNO2.
The pH of a solution is 14 minus the pOH, which in this case is 14 minus 5.559.


 HNO2  OH 
Kb 
 NO 2 


x2
Kb 
 0.35  x 
x2
Kb 
0.35
x 2  0.35 K b
 OH    x  0.35 K b



 OH    0.35 2.17  10 11


 OH    7.595  10 12  2.76  10 6 M




pOH   log 2.76  10 6  5.559
pH  14.000  5.559  8.441
Find the pH of a 0.35 M solution of NaNO2.
Which comes out to 8.441 .



 HNO2  OH 
Kb 
 NO 2 


x2
Kb 
 0.35  x 
x2
Kb 
0.35
x 2  0.35 K b
 OH    x  0.35 K b



 OH    0.35 2.17  10 11



 OH    7.595  10 12  2.76  10 6 M




pOH   log 2.76  10 6  5.559
pH  14.000  5.559  8.44
Find the pH of a 0.35 M solution of NaNO2.
Both the given concentration of 0.35 molar and the Ka for HNO2 we used from the acid table have 2
significant figures. So we round our final pH to 2 significant figures, or 2 decimal places, so its 8.44


 HNO2  OH 
Kb 
 NO 2 


x2
Kb 
 0.35  x 
x2
Kb 
0.35
x 2  0.35 K b
 OH    x  0.35 K b



 OH    0.35 2.17  10 11



 OH    7.595  10 12  2.76  10 6 M




pOH   log 2.76  10 6  5.559
pH  14.000  5.559
pH  8.44
Find the pH of a 0.35 M solution of NaNO2.
So now we have a final answer for the question. The pH of a 0.35 molar solution of
NaNO2 is 8.44 .


 HNO2  OH 
Kb 
 NO 2 


x2
Kb 
 0.35  x 
x2
Kb 
0.35
K b(NO )  2.17  10 11
2
x 2  0.35 K b
Low
 OH    x  0.35 K b



 OH    0.35 2.17  10 11



 OH    7.595  10 12  2.76  10 6 M




pOH   log 2.76  10 6  5.559
pH  14.000  5.559
pH  8.44
Find the pH of a 0.35 M solution of NaNO2.
This is reasonable because NO2 minus is a weak base, so we would expect its pH to
be above 7, but not really high because its Kb value is quite low at 2.17 × 10-11.