Here we’ll go through the steps necessary to find the pH of a weak base with a given...
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Transcript of Here we’ll go through the steps necessary to find the pH of a weak base with a given...
Here we’ll go through the steps necessary to find the pH of a weak base with a given concentration.
Finding the pH of a Weak
Base
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.
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.
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.
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.
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–].
Finally we convert hydronium or hydroxide concentration to pH.
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.
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.Dissociate
Which are Na+ and NO2 minus ions.
Find the pH of a 0.35 M solution of NaNO2.Dissociate
(aq) 2(aq)2(aq)NaNO Na NO
Neutral Spectato
r
WeakBase
Na+ is an alkali metal cation, so it is a neutral spectator.
Find the pH of a 0.35 M solution of NaNO2.Dissociate
(aq2(aq) 2(aq))NaNO NONa
Neutral Spectato
r
WeakBase
And we can discard it.
Find the pH of a 0.35 M solution of NaNO2.Dissociate
(aq2(aq) 2(aq))NaNO NONa
Neutral Spectato
r
WeakBase
In order to determine what NO2 minus acts as, we look for it on the acid table.
Find the pH of a 0.35 M solution of NaNO2.Dissociate
2(aq) (aq) 2(aq)NaNO NONa
?
It’s at this location on the right side of the table, so it’s a weak base.
Weak Base
Because NO2 minus is a weak base
2(aq) (aq) 2(aq)NaNO Na NO
WeakBase
Find the pH of a 0.35 M solution of NaNO2.
We will need to find the value of its Kb
Find the pH of a 0.35 M solution of NaNO2.
2(aq) (aq) 2(aq)NaNO NONa
WeakBase
2
2
w
a(conjugate acid)
w
a(H
b(N
NO )
141
O )
14
1.00 102.17 10
4.6 10
K
K
K
K
K
Remember the formula we can use for Kb of NO2 minus is Kb = Kw over the Ka of its conjugate acid.
Find the pH of a 0.35 M solution of NaNO2.
2
2
b(NO )
w
a(H
w
a(conjugate acid
NO )
141
4
)
11.00 102.17 10
4.6 10
KK
K
K
K
Looking at the table, we see that the conjugate acid of NO2 minus is HNO2.
The conjugate acid of NO2
– is HNO2
2
2
2
wb(NO )
a(HNO
wb
)
NOcon
( )a( )
14
jugate acid
114
1.00 102.17 10
4.6 10
KK
KK
K
K
So the Kb of NO2 minus is Kw over the Ka of HNO2.
Find the pH of a 0.35 M solution of NaNO2.
The Ka for HNO2 is shown on the table here, and its 4.6 × 10-4.
Ka of HNO2
2
2
2
w
wb(NO )
a(conjugate acid)
b(NO )a(HNO )
11
41
41.00 10
4.6 12.17 0
01
KK
K
KK
K
Now we can substitute. We know that Kw is 1.00 × 10-14,
Find the pH of a 0.35 M solution of NaNO2.
2
2
2
wb(NO )
a(conjugate acid)
wb(NO )
a(HNO
1
)
411
4
1.00
4
102.17
.10
6 10
KK
K
KK
K
And the Ka of HNO2 is 4.6 × 10-4.
Find the pH of a 0.35 M solution of NaNO2.
2
2
2
wb(NO )
a(conjugate acid)
w
a(HNO )
11b(NO )
14
4
1.00 10
4.6 117
02. 10K
KK
K
K
K
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.
Find the pH of a 0.35 M solution of NaNO2.
2
2
2
wb(NO )
a(conjugate acid)
w
a(HNO )
11b(NO )
14
4
1.00 10
4.6 117
02. 10K
KK
K
K
K
So we’ll make a note up here that the Kb of NO2 minus is 2.17 times 10-11.
Find the pH of a 0.35 M solution of NaNO2.
2
11b(NO )
2.17 10K
Because NO2 minus is a weak base, we can write its hydrolysis equilibrium equation.
Find the pH of a 0.35 M solution of NaNO2.
2
11b(NO )
2.17 10K
2(aq) 2 ( ) 2(aq) (aq)NO H O HNO OH l
Weak Base
Hydrolysis means we add it to water.
Find the pH of a 0.35 M solution of NaNO2.
2
11b(NO )
2.17 10K
2 ( ) 2(aq) (aq)2(aq) H O H NO NO H Ol
Because it’s a base, it will accept a proton from water.
Find the pH of a 0.35 M solution of NaNO2.
2
11b(NO )
2.17 10K
2(aq) 2 ( ) 2(aq) (aq)HNO NO H H O Ol
H+
Weak Base
Forming HNO2
Find the pH of a 0.35 M solution of NaNO2.
2
11b(NO )
2.17 10K
2(aq) 2 ( ) 2(aq) (aq)HNO NO H H O Ol
H+
And OH minus.
Find the pH of a 0.35 M solution of NaNO2.
2
11b(NO )
2.17 10K
2(aq) 2 ( ) (a2(aq) q) NO H O HNO OHl
H+
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.
Find the pH of a 0.35 M solution of NaNO2.
2
11b(NO )
2.17 10K
2(aq) 2 ( ) (a2(aq) q) NO H O HNO OHl
?
So we draw an ICE table underneath this equilibrium equation.
Find the pH of a 0.35 M solution of NaNO2.
2
11b(NO )
2.17 10K
[I][C]
[E]
2(aq) 2 ( ) 2(aq) (aq) NO H O HNO OHl
Water is a liquid so we don’t include it in our calculations. We’ll colour the column below water blue.
Find the pH of a 0.35 M solution of NaNO2.
2
11b(NO )
2.17 10K
[I][C]
[E]
2(aq) 2 ( ) 2(aq) (aq) NO H O HNO OHl
Now we’ll fill in what we can in the initial concentration row.
Find the pH of a 0.35 M solution of NaNO2.
2
11b(NO )
2.17 10K
[I] 0.35 0 0[C] –x +x +x
[E] 0.35–x x x
2(aq) 2 ( ) 2(aq) (aq) NO H O HNO OHl
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.
Find the pH of a 0.35 M solution of NaNO2.
2
11b(NO )
2.17 10K
[I] 0.35 0 0[C] –x +x +x
[E] 0.35–x x x
2(aq) 2 ( ) 2(aq) (aq) NO H O HNO OHl
Before hydrolysis occurs, we can say that the concentrations of HNO2 and OH minus are both zero.
Find the pH of a 0.35 M solution of NaNO2.
2
11b(NO )
2.17 10K
[I] 0.35 0 0[C] –x +x +x
[E] 0.35–x x x
2(aq) 2 ( ) 2(aq) (aq) NO H O HNO OHl
Now, we’ll fill in the change in concentration row.
Find the pH of a 0.35 M solution of NaNO2.
2
11b(NO )
2.17 10K
[I] 0.35 0 0[C] –x +x +x
[E] 0.35–x x x
2(aq) 2 ( ) 2(aq) (aq) NO H O HNO OHl
Because the concentrations of the products are zero, in order to compensate…
Find the pH of a 0.35 M solution of NaNO2.
2
11b(NO )
2.17 10K
[I] 0.35 0 0[C] –x +x +x
[E] 0.35–x x x
2(aq) 2 ( ) 2(aq) (aq) NO H O HNO OHl
The reaction will move to the right.
Find the pH of a 0.35 M solution of NaNO2.
2
11b(NO )
2.17 10K
[I] 0.35 0 0[C] –x +x +x
[E] 0.35–x x x
2(aq) 2 ( ) 2(aq) (aq) NO H O HNO OHl
Moves to the Right
So the concentrations of HNO2 and OH minus will both increase
Find the pH of a 0.35 M solution of NaNO2.
2
11b(NO )
2.17 10K
[I] 0.35 0 0[C] –x +x +x
[E] 0.35–x x x
2(aq) 2 ( ) 2(aq) (aq) NO H O HNO OHl
Moves to the Right
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
Find the pH of a 0.35 M solution of NaNO2.
2
11b(NO )
2.17 10K
[I] 0.35 0 0[C] –x +x +x
[E] 0.35–x x x
2(aq) 2 ( ) 2(aq) (aq) NO H O HNO OHl
Moves to the Right
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.
Find the pH of a 0.35 M solution of NaNO2.
2
11b(NO )
2.17 10K
[I] 0.35 0 0[C] –x +x +x
[E] 0.35–x x x
2(aq) 2 ( ) 2(aq) (aq) NO H O HNO OHl
Moves to the Right
Like the HNO2 and the OH minus, the coefficient on NO2 minus is 1,
Find the pH of a 0.35 M solution of NaNO2.
2
11b(NO )
2.17 10K
[I] 0.35 0 0[C] –x +x +x
[E] 0.35–x x x
2(aq) 2 ( ) 2(aq) (aq) NO H O HNO OHl
Moves to the Right
So we can say that the concentration of NO2 minus goes down by x.
Find the pH of a 0.35 M solution of NaNO2.
2
11b(NO )
2.17 10K
[I] 0.35 0 0[C] –x +x +x
[E] 0.35–x x x
2(aq) 2 ( ) 2(aq) (aq) NO H O HNO OHl
Moves to the Right
Now we can fill in the equilibrium concentration row.
Find the pH of a 0.35 M solution of NaNO2.
2
11b(NO )
2.17 10K
[I] 0.35 0 0[C] –x +x +x
[E] 0.35–x x x
2(aq) 2 ( ) 2(aq) (aq) NO H O HNO OHl
We’ll start with the OH minus. It will be 0 plus x
Find the pH of a 0.35 M solution of NaNO2.
2
11b(NO )
2.17 10K
[I] 0.35 0 0[C] –x +x +x
[E] 0.35–x x x
2(aq) 2 ( ) 2(aq) (aq) NO H O HNO OHl
Which is equal to x
Find the pH of a 0.35 M solution of NaNO2.
2
11b(NO )
2.17 10K
[I] 0.35 0 0[C] –x +x +x
[E] 0.35–x x x
2(aq) 2 ( ) 2(aq) (aq) NO H O HNO OHl
Similarly, the equilibrium concentration of HNO2 will be 0 + x
Find the pH of a 0.35 M solution of NaNO2.
2
11b(NO )
2.17 10K
[I] 0.35 0 0[C] –x +x +x
[E] 0.35–x x x
2(aq) 2 ( ) 2(aq) (aq) NO H O HNO OHl
Which is equal to x
Find the pH of a 0.35 M solution of NaNO2.
2
11b(NO )
2.17 10K
[I] 0.35 0 0[C] –x +x +x
[E] 0.35–x x x
2(aq) 2 ( ) 2(aq) (aq) NO H O HNO OHl
Now, we’ll look at the NO2 minus. It started out as 0.35 molar
Find the pH of a 0.35 M solution of NaNO2.
2
11b(NO )
2.17 10K
[I] 0.35 0 0[C] –x +x +x
[E] 0.35–x x x
2(aq) 2 ( ) 2(aq) (aq) NO H O HNO OHl
And it went down by x,
Find the pH of a 0.35 M solution of NaNO2.
2
11b(NO )
2.17 10K
[I] 0.35 0 0[C] –x +x +x
[E] 0.35–x x x
2(aq) 2 ( ) 2(aq) (aq) NO H O HNO OHl
So its equilibrium concentration is 035 minus x
Find the pH of a 0.35 M solution of NaNO2.
2
11b(NO )
2.17 10K
[I] 0.35 0 0[C] –x +x +x
[E] 0.35–x x x
2(aq) 2 ( ) 2(aq) (aq) NO H O HNO OHl
So now we have the equilibrium concentrations of all species.
Find the pH of a 0.35 M solution of NaNO2.
2
11b(NO )
2.17 10K
[I] 0.35 0 0[C] –x +x +x
[E] 0.35–x x x
2(aq) 2 ( ) 2(aq) (aq) NO H O HNO OHl
Notice that the equilibrium concentration of OH minus is x.
Find the pH of a 0.35 M solution of NaNO2.
2
11b(NO )
2.17 10K
[I] 0.35 0 0[C] –x +x +x
[E] 0.35–x x x
2(aq) 2 ( ) 2(aq) (aq) NO H O HNO OHl
We’ll make a note of that up here.
Find the pH of a 0.35 M solution of NaNO2.
2
11b(NO )
2.17 10K
[I] 0.35 0 0[C] –x +x +x
[E] 0.35–x x x
2(aq) 2 ( ) 2(aq) (aq) NO H O HNO OHl
OH x
At this point, we need to solve for the value of x.
Find the pH of a 0.35 M solution of NaNO2.
2
11b(NO )
2.17 10K
[I] 0.35 0 0[C] –x +x +x
[E] 0.35–x x x
2(aq) 2 ( ) 2(aq) (aq) NO H O HNO OHl
OH x
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.
Find the pH of a 0.35 M solution of NaNO2.
2
11b(NO )
2.17 10K OH x
2b
2
2
b
2
b
HNO OH
N
x
0.35 x
x
0.3
O
5
K
K
K
Remember, because we’re dealing with the hydrolysis of a weak BASE, the expression is called Kb rather than Ka.
Find the pH of a 0.35 M solution of NaNO2.
2
11b(NO )
2.17 10K OH x
2
2
b
2
b
b2
N
HN
x
0.35
O O
.
H
0 3
O
x
x
5
K
K
K
Weak Base
Now we’ll insert equilibrium concentrations into the Kb expression in order to solve for x.
Find the pH of a 0.35 M solution of NaNO2.
2
11b(NO )
2.17 10K OH x
2b
2
2
b
2
b
HNO OH
N
x
0.35 x
x
0.3
O
5
K
K
K
The concentration of HNO2 and OH minus are both equal to x,
Find the pH of a 0.35 M solution of NaNO2.
2
11b(NO )
2.17 10K OH x
b2
2
2
b
b
2
N
x
HNO OH
O
0.35
x
x
0.35
K
K
K
so their product in the Kb expression is x times x, or x squared
Find the pH of a 0.35 M solution of NaNO2.
2
11b(NO )
2.17 10K OH x
b2
2
2
b
b
2
N
x
HNO OH
O
0.35
x
x
0.35
K
K
K
And the equilibrium concentration of NO2 minus is 0.35 minus x,
Find the pH of a 0.35 M solution of NaNO2.
2
11b(NO )
2.17 10K OH x
2
2
2
b
b
2b
N
HN
0.35
x
0.3
O OH
x
5
x
OK
K
K
so we’ll substitute that in here for the concentration of NO2 minus.
Find the pH of a 0.35 M solution of NaNO2.
2
11b(NO )
2.17 10K OH x
2
2
2
b
b
2b
N
HN
0.35
x
0.3
O OH
x
5
x
OK
K
K
In order to avoid a quadratic equation, we assume that 0.35 minus x is almost equal to 0.35.
Find the pH of a 0.35 M solution of NaNO2.
2
11b(NO )
2.17 10K OH x
2b
2
b
2
2
b
HNO OH
N
x
0
0.3 x
O
x
5
.35
K
K
K
Assume0.35–x 0.35
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
Find the pH of a 0.35 M solution of NaNO2.
2
11b(NO )
2.17 10K OH x
2b
2
b
2
2
b
HNO OH
N
x
0
0.3 x
O
x
5
.35
K
K
K
Assume0.35–x 0.35
Using this assumption, we take the x out of the denominator
Find the pH of a 0.35 M solution of NaNO2.
2
11b(NO )
2.17 10K OH x
2b
2
b
2
2
b
HNO OH
N
x
0
0.3 x
O
x
5
.35
K
K
K
Assume0.35–x 0.35
And we get that Kb is approximately equal to x squared divided by 0.35.
Find the pH of a 0.35 M solution of NaNO2.
2
11b(NO )
2.17 10K OH x
2b
2
b
2
2
b
HNO OH
N
x
0
0.3 x
O
x
5
.35
K
K
K
Rearranging this equation gives us x squared = 0.35 times Kb
Find the pH of a 0.35 M solution of NaNO2.
2
11b(NO )
2.17 10K OH x
2b
2
b
2
2
b
HNO OH
N
x
0
O
x
0.35 x
.35
K
K
K
b
11
12 6
2
6
b
OH 0.35
OH 0.35 2.17 10
OH 7.595 10 2.76 10 M
pOH log 2.76 10 5.559
14.000 5.559 8.4
x 0.35
4
x K
pH
K
Taking the square root of both sides gives us x = the square root of 0.35 times kb
Find the pH of a 0.35 M solution of NaNO2.
2
11b(NO )
2.17 10K OH x
2b
2
2
b
2
b
HNO OH
NO
x
0.35 x
x
0.35
K
K
K
11
12 6
6
b
2b
OH
OH 0.35 2.17 10
OH 7.595 10 2.76 10 M
pOH log 2.76 10 5.559
14.000 5.559 8
x 0.
x 0.3
. 4
35
5
4
K
pH
K
Take square root of
both sides
Remember from our ice table, that x is equal to the hydroxide ion concentration at equilibrium.
Find the pH of a 0.35 M solution of NaNO2.
2
11b(NO )
2.17 10K OH x
2b
2
2
b
2
b
HNO OH
NO
x
0.35 x
x
0.35
K
K
K
2b
b
11
12 6
6
x 0.35
x 0.35
OH 0.35 2.17 10
OH 7.595 10 2.76 10 M
pOH log 2.76 10 5.559
14.000 5.559
OH
8.44
K
K
pH
So we can say that the hydroxide ion concentration is equal to x, which is equal to the square root of 0.35 kb
Find the pH of a 0.35 M solution of NaNO2.
2
11b(NO )
2.17 10K OH x
2b
2
2
b
2
b
HNO OH
NO
x
0.35 x
x
0.35
K
K
K
2b
11
12 6
6
b
x 0.35
OH 0.35 2.17 10
OH 7.595 10 2.76 10 M
pOH log 2.76 10 5.559
14.000 5.5
OH x 0.
59 8.
5
44
3 K
K
pH
Or more simply, [OH-] is equal to the square root of 0.35 kb
Find the pH of a 0.35 M solution of NaNO2.
2
11b(NO )
2.17 10K OH x
2b
2
2
b
2
b
HNO OH
NO
x
0.35 x
x
0.35
K
K
K
2b
11
12 6
6
b
x 0.35
OH 0.35 2.17 10
OH 7.595 10 2.76 10 M
pOH log 2.76 10 5.559
14.000 5.5
OH x 0.
59 8.
5
44
3 K
K
pH
Now we’ll substitute 2.17 × 10-11 in for the value of Kb in the equation
Find the pH of a 0.35 M solution of NaNO2.
2
11b(NO )
2.17 10K
2b
2
2
b
2
b
HNO OH
NO
x
0.35 x
x
0.35
K
K
K
11
2b
12 6
6
b
x 0.35
OH x 0.35
OH 0.35
OH 7.595 10 2.76 10 M
pOH log 2.76 10 5.559
14.000 5.559 8.
2.17 1
44
0
K
pH
K
0.35 times 2.17 × 10-11 is equal to 7.595 × 10-12.
Find the pH of a 0.35 M solution of NaNO2.
2
11b(NO )
2.17 10K
2b
2
2
b
2
b
HNO OH
NO
x
0.35 x
x
0.35
K
K
K
11
12
2b
b
6
6
x 0.35
OH x 0.35
OH
O 7.59H 2.76 10 M
pOH log 2.76 10 5.559
14.000 5.559 8.44
0.35 2.
5 1
17 10
0
K
K
pH
Taking the square root of 7.595 × 10-12 gives us 2.76 × 10-6.
2b
2
2
b
2
b
HNO OH
NO
x
0.35 x
x
0.35
K
K
K
2b
b
11
61
6
2
x 0.35
OH x 0.35
OH 0.35 2.17 10
OH
pOH log 2.76 10 5.559
14.000 5.559 8
7.595 2.710
.44
6 10 M
K
K
pH
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.
Find the pH of a 0.35 M solution of NaNO2.
2b
2
2
b
2
b
HNO OH
NO
x
0.35 x
x
0.35
K
K
K
2b
b
11
12 6
6
x 0.35
OH x 0.35
OH 0.35 2.17 10
7.595 10 2.76 10
pOH log 2.76 10 5.559
14.000 5.559 8.
MH
4
O
4
K
K
pH
pOH is the negative log of the hydroxide ion concentration, which is the negative log of 2.76 × 10-6.
Find the pH of a 0.35 M solution of NaNO2.
2b
2
2
b
2
b
HNO OH
NO
x
0.35 x
x
0.35
K
K
K
6
2b
b
11
1 62
x 0.35
OH x 0.35
OH 0.35 2.17 10
OH 7.595 10 M
5.559
14.000 5.559 8.4
pOH log 2.7
2.
6
76 10
10
4
K
K
pH
And that comes out to 5.559. We’ll stick with 3 significant figures here and round off to 2 in the last step.
Find the pH of a 0.35 M solution of NaNO2.
2b
2
2
b
2
b
HNO OH
NO
x
0.35 x
x
0.35
K
K
K
6
2b
b
11
1
6
2
x 0.35
OH x 0.35
OH 0.35 2.17 10
OH 7.595 10 M
14.000 5.559 8.4
2.76 10
log 2.76pOH 90 5. 51 5
4
K
K
pH
The pH of a solution is 14 minus the pOH, which in this case is 14 minus 5.559.
Find the pH of a 0.35 M solution of NaNO2.
2b
2
2
b
2
b
HNO OH
NO
x
0.35 x
x
0.35
K
K
K
2b
b
11
12 6
6
x 0.35
OH x 0.35
OH 0.35 2.17 10
OH 7.595 10 2.76 10 M
log 2.76 10
8.4
pOH 5.559
pH 14.000 5.559 4
K
K
Which comes out to 8.441 .
Find the pH of a 0.35 M solution of NaNO2.
2b
2
2
b
2
b
HNO OH
NO
x
0.35 x
x
0.35
K
K
K
2b
b
11
12 6
6
14.000 5
x 0.35
OH x 0.35
OH 0.35 2.17 10
OH 7.595 10 2.76 10 M
pOH log 2
pH 8.559
.76 10 5.55
.4 1
9
4
K
K
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
Find the pH of a 0.35 M solution of NaNO2.
2b
2
2
b
2
b
HNO OH
NO
x
0.35 x
x
0.35
K
K
K
2b
b
11
12 6
6
pH 4
x 0.35
OH x 0.35
OH 0.35 2.17 10
OH 7.595 10 2.76 10 M
pOH log 2.76 10 5.559
14.000 5.559 8 4.
K
K
So now we have a final answer for the question. The pH of a 0.35 molar solution of NaNO2 is 8.44 .
Find the pH of a 0.35 M solution of NaNO2.
2b
2
2
b
2
b
HNO OH
NO
x
0.35 x
x
0.35
K
K
K
2b
b
11
12 6
6
pH 8.4
x 0.35
OH x 0.35
OH 0.35 2.17 10
OH 7.595 10 2.76 10 M
pOH log 2.76 10 5.559
14.000 5.559 4
K
K
pH
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.
Find the pH of a 0.35 M solution of NaNO2.
2
11b(NO )
2.17 10K
2b
2
2
b
2
b
HNO OH
NO
x
0.35 x
x
0.35
K
K
K
2b
b
11
12 6
6
pH 8.4
x 0.35
OH x 0.35
OH 0.35 2.17 10
OH 7.595 10 2.76 10 M
pOH log 2.76 10 5.559
14.000 5.559 4
K
K
pH
Low