pH and Kb to
Initial
Concentration
of a Base
Here we’ll show you how to find the initial concentration of a base, given the pH
and Kb of a solution of the base.
The weak base dimethylamine,
(CH3)2NH, has a Kb value of 5.1 × 10–4.
We’re given that the weak base dimethylamine, (CH3)2NH, has a Kb value of
5.1 × 10–4.
The weak base dimethylamine,
(CH3)2NH, has a Kb value of 5.1 × 10–4.
A particular solution of dimethlyamine
has a pH of 11.9435.
We’re told that a particular solution of dimethlyamine is found to have a pH of
11.9435.
The weak base dimethylamine,
(CH3)2NH, has a Kb value of 5.1 × 10–4.
A particular solution of dimethlyamine
has a pH of 11.9435.
What was the initial [(CH3)2NH]?
And we’re asked to determine the initial concentration of dimethylamine in that
solution.
Steps we can use to solve this problem:
The weak base dimethylamine, (CH3)2NH, has a Kb value of
5.1 × 10–4. A particular solution of dimethlyamine has a pH of
11.9435. What was the initial [(CH3)2NH]?
Here are the general steps we can take to solve this problem.
Steps we can use to solve this problem:
1. Convert pH  pOH  [OH–]
The weak base dimethylamine, (CH3)2NH, has a Kb value of
5.1 × 10–4. A particular solution of dimethlyamine has a pH of
11.9435. What was the initial [(CH3)2NH]?
We will need the hydroxide ion concentration and we are given the pH, so we can
start by converting pH to pOH and then to hydroxide ion concentration.
Steps we can use to solve this problem:
1. Convert pH  pOH  [OH–]
2. Write equilibrium equation for ionization
of (CH3)2NH
The weak base dimethylamine, (CH3)2NH, has a Kb value of
5.1 × 10–4. A particular solution of dimethlyamine has a pH of
11.9435. What was the initial [(CH3)2NH]?
Like all problems involving weak acids or bases, we write out the equilibrium equation. Because
dimethylamine is a Neutral weak base, the reaction is called ionization rather then hydrolysis.
Steps we can use to solve this problem:
1. Convert pH  pOH  [OH–]
2. Write equilibrium equation for ionization
of (CH3)2NH
3. Create an ICE table
The weak base dimethylamine, (CH3)2NH, has a Kb value of
5.1 × 10–4. A particular solution of dimethlyamine has a pH of
11.9435. What was the initial [(CH3)2NH]?
Because we’re doing calculations involving a Weak base, we create an ICE table
below the equilibrium equation.
Steps we can use to solve this problem:
1. Convert pH  pOH  [OH–]
2. Write equilibrium equation for ionization
of (CH3)2NH
3. Create an ICE table
4. Use the ICE table and Kb expression to
determine the initial [(CH3)2NH]
The weak base dimethylamine, (CH3)2NH, has a Kb value of
5.1 × 10–4. A particular solution of dimethlyamine has a pH of
11.9435. What was the initial [(CH3)2NH]?
We can then use this ICE table, along with the Kb expression for dimethylamine, to
determine the Initial concentration of this base.
Convert pH  pOH  [OH–]
The weak base dimethylamine, (CH3)2NH, has a Kb value of
5.1 × 10–4. A particular solution of dimethlyamine has a pH of
11.9435. What was the initial [(CH3)2NH]?
We’re given the pH and we need the hydroxide ion concentration, so we start by
converting pH to pOH , and then to hydroxide ion concentration.
pH  11.9435
pOH  14.000  pH
pOH  14.000  11.9435  2.0565
Convert pH  pOH  [OH–]
 OH    10  pOH  10 2.0565


 OH    0.00878 M


The weak base dimethylamine, (CH3)2NH, has a Kb value of
5.1 × 10–4. A particular solution of dimethlyamine has a pH of
11.9435. What was the initial [(CH3)2NH]?
The pH is 11.9435
pH  11.9435
pOH  14.000  pH
pOH  14.000  11.9435  2.0565
Convert pH  pOH  [OH–]
 OH    10  pOH  10 2.0565


 OH    0.00878 M


The weak base dimethylamine, (CH3)2NH, has a Kb value of
5.1 × 10–4. A particular solution of dimethlyamine has a pH of
11.9435. What was the initial [(CH3)2NH]?
The pOH is 14 minus the pH.
pH  11.9435
pOH  14.000  pH
pOH  14.000  11.9435  2.0565
Convert pH  pOH  [OH–]
 OH    10  pOH  10 2.0565


 OH    0.00878 M


The weak base dimethylamine, (CH3)2NH, has a Kb value of
5.1 × 10–4. A particular solution of dimethlyamine has a pH of
11.9435. What was the initial [(CH3)2NH]?
Which is 14 minus 11.9435. Don’t worry about significant figures at this point.
We’ll use these numbers as they are and round off later.
pH  11.9435
pOH  14.000  pH
pOH  14.000  11.9435  2.0565
Convert pH  pOH  [OH–]
 OH    10  pOH  10 2.0565


 OH    0.00878 M


The weak base dimethylamine, (CH3)2NH, has a Kb value of
5.1 × 10–4. A particular solution of dimethlyamine has a pH of
11.9435. What was the initial [(CH3)2NH]?
14 minus 11.9435 comes out to 2.0565
pH  11.9435
pOH  14.000  pH
pOH  14.000  11.9435  2.0565
Convert pH  pOH  [OH–]
 OH    10  pOH  10 2.0565


 OH    0.00878 M


The weak base dimethylamine, (CH3)2NH, has a Kb value of
5.1 × 10–4. A particular solution of dimethlyamine has a pH of
11.9435. What was the initial [(CH3)2NH]?
The concentration of hydroxide ion is 10 to the negative pOH
pH  11.9435
pOH  14.000  pH
pOH  14.000  11.9435  2.0565
Convert pH  pOH  [OH–]
 OH    10  pOH  10 2.0565


 OH    0.00878 M


The weak base dimethylamine, (CH3)2NH, has a Kb value of
5.1 × 10–4. A particular solution of dimethlyamine has a pH of
11.9435. What was the initial [(CH3)2NH]?
Which is 10 to the negative 2.0565
pH  11.9435
pOH  14.000  pH
pOH  14.000  11.9435  2.0565
Convert pH  pOH  [OH–]
 OH    10  pOH  10 2.0565


 OH    0.00878 M


The weak base dimethylamine, (CH3)2NH, has a Kb value of
5.1 × 10–4. A particular solution of dimethlyamine has a pH of
11.9435. What was the initial [(CH3)2NH]?
This comes out to 0.00878 . Because the Kb has 2 significant figures, the final answer to our problem will
need to have 2 significant figures, but we’ll use this number for now, expressed as 3 significant figures,
pH  11.9435
pOH  14.000  pH
pOH  14.000  11.9435  2.0565
Convert pH  pOH  [OH–]
 OH    10  pOH  10 2.0565


 OH    0.00878 M


The weak base dimethylamine, (CH3)2NH, has a Kb value of
5.1 × 10–4. A particular solution of dimethlyamine has a pH of
11.9435. What was the initial [(CH3)2NH]?
So we can now state that the hydroxide ion concentration is 0.00878 molar.
[OH ] = 0.00878 M
pH  11.9435
pOH  14.000  pH
pOH  14.000  11.9435  2.0565
–
 OH    10  pOH  10 2.0565


 OH    0.00878 M


The weak base dimethylamine, (CH3)2NH, has a Kb value of
5.1 × 10–4. A particular solution of dimethlyamine has a pH of
11.9435. What was the initial [(CH3)2NH]?
And we’ll make a note of that up here.
[OH–] = 0.00878 M
(CH 3 ) 2 NH(aq)  H 2O( l )


(CH 3 ) 2 NH 2(aq)
 OH(aq)
The weak base dimethylamine, (CH3)2NH, has a Kb value of
5.1 × 10–4. A particular solution of dimethlyamine has a pH of
11.9435. What was the initial [(CH3)2NH]?
Now, we’ll write the equilibrium equation for the ionization of dimethylamine. We
start by writing (CH3)2NH(aq)
[OH–] = 0.00878 M
(CH 3 ) 2 NH(aq)  H 2O( l )


(CH 3 ) 2 NH 2(aq

OH
)
( aq )
The weak base dimethylamine, (CH3)2NH, has a Kb value of
5.1 × 10–4. A particular solution of dimethlyamine has a pH of
11.9435. What was the initial [(CH3)2NH]?
In its ionization equation, it’s added to water.
[OH–] = 0.00878 M
H+
(CH 3 ) 2 NH(aq)  H 2O( l )


(CH 3 ) 2 NH 2(aq

OH
)
( aq )
The weak base dimethylamine, (CH3)2NH, has a Kb value of
5.1 × 10–4. A particular solution of dimethlyamine has a pH of
11.9435. What was the initial [(CH3)2NH]?
The base (CH3)2NH will accept a proton from water.
H+
(CH 3 ) 2 NH(aq)  H 2O( l )
[OH–] = 0.00878 M


(CH 3 ) 2 NH 2(

OH
aq)
(aq )
The weak base dimethylamine, (CH3)2NH, has a Kb value of
5.1 × 10–4. A particular solution of dimethlyamine has a pH of
11.9435. What was the initial [(CH3)2NH]?
And produce the conjugate acid (CH3)2NH2+
H+
(CH 3 ) 2 NH(aq)  H 2O( l )
[OH–] = 0.00878 M


(CH 3 ) 2 NH 2(aq

O
H
)
(aq )
The weak base dimethylamine, (CH3)2NH, has a Kb value of
5.1 × 10–4. A particular solution of dimethlyamine has a pH of
11.9435. What was the initial [(CH3)2NH]?
And a hydroxide ion.
[OH–] = 0.00878 M
(CH 3 ) 2 NH(aq)  H 2O( l )

(CH 3 ) 2 NH 2(aq)
 OH(aq)
Weak
Base
The weak base dimethylamine, (CH3)2NH, has a Kb value of
5.1 × 10–4. A particular solution of dimethlyamine has a pH of
11.9435. What was the initial [(CH3)2NH]?
Because (CH3)2NH is a Weak base, we add an ICE table under this equation.
[OH–] = 0.00878 M
(CH 3 ) 2 NH(aq)  H 2O( l )


(CH 3 ) 2 NH 2(aq)
 OH(aq)
[I]
[C]
[E]
The weak base dimethylamine, (CH3)2NH, has a Kb value of
5.1 × 10–4. A particular solution of dimethlyamine has a pH of
11.9435. What was the initial [(CH3)2NH]?
Like this
[OH–] = 0.00878 M
(CH3 )2 NH(aq)  H2O( l )


(CH3 ) 2 NH2(aq)
 OH(aq)
[I]
[C]
[E]
The weak base dimethylamine, (CH3)2NH, has a Kb value of
5.1 × 10–4. A particular solution of dimethlyamine has a pH of
11.9435. What was the initial [(CH3)2NH]?
As always, we’ll leave out water because it’s a liquid.
[OH–] = 0.00878 M
(CH3 )2 NH(aq)  H2O( l )


(CH3 ) 2 NH2(aq)
 OH(aq)
[I]
[C]
[E]
The weak base dimethylamine, (CH3)2NH, has a Kb value of
5.1 × 10–4. A particular solution of dimethlyamine has a pH of
11.9435. What was the initial [(CH3)2NH]?
Now we’re ready to fill in the row for initial concentrations.
[OH–] = 0.00878 M
(CH3 )2 NH(aq)  H2O( l )
[I]
[C]
[E]


(CH3 ) 2 NH2(aq)
 OH(aq)
?
The weak base dimethylamine, (CH3)2NH, has a Kb value of
5.1 × 10–4. A particular solution of dimethlyamine has a pH of
11.9435. What was the initial [(CH3)2NH]?
What we’re actually trying to find out in this problem is the initial concentration of
dimethylamine, so that is our unknown.
[OH–] = 0.00878 M
(CH3 )2 NH(aq)  H2O( l )
[I]
[C]
[E]


(CH3 ) 2 NH2(aq)
 OH(aq)
x
The weak base dimethylamine, (CH3)2NH, has a Kb value of
5.1 × 10–4. A particular solution of dimethlyamine has a pH of
11.9435. What was the initial [(CH3)2NH]?
So we’ll let x equal the initial concentration of dimethylamine.
[OH–] = 0.00878 M
(CH3 )2 NH(aq)  H2O( l )
[I]
[C]
[E]
x


(CH3 ) 2 NH2(aq)
 OH(aq)
0
0
The weak base dimethylamine, (CH3)2NH, has a Kb value of
5.1 × 10–4. A particular solution of dimethlyamine has a pH of
11.9435. What was the initial [(CH3)2NH]?
And the initial concentrations of (CH3)2NH2+ and OH minus are both equal to zero.
We have now finished filling in the row for initial concentrations.
[OH–] = 0.00878 M
(CH3 )2 NH(aq)  H2O( l )
[I]
[C]
[E]
x


(CH3 ) 2 NH2(aq)
 OH(aq)
0
0
0.00878
The weak base dimethylamine, (CH3)2NH, has a Kb value of
5.1 × 10–4. A particular solution of dimethlyamine has a pH of
11.9435. What was the initial [(CH3)2NH]?
We know the equilibrium concentration of OH minus is 0.00878 molar, so we’ll add
it down here in the equilibrium concentration row.
(CH3 )2 NH(aq)  H2O( l )
[I]
[C]
[E]


(CH3 ) 2 NH2(aq)
 OH(aq)
x
0
0
0.00878
The weak base dimethylamine, (CH3)2NH, has a Kb value of
5.1 × 10–4. A particular solution of dimethlyamine has a pH of
11.9435. What was the initial [(CH3)2NH]?
The hydroxide ion concentration started out as 0…
(CH3 )2 NH(aq)  H2O( l )
[I]
[C]
[E]
x


(CH3 ) 2 NH2(aq)
 OH(aq)
0
0
0.00878
The weak base dimethylamine, (CH3)2NH, has a Kb value of
5.1 × 10–4. A particular solution of dimethlyamine has a pH of
11.9435. What was the initial [(CH3)2NH]?
And ended up as 0.00878 molar
(CH3 )2 NH(aq)  H2O( l )
[I]
[C]
[E]
x


(CH3 ) 2 NH2(aq)
 OH(aq)
0
0
+0.00878
0.00878
The weak base dimethylamine, (CH3)2NH, has a Kb value of
5.1 × 10–4. A particular solution of dimethlyamine has a pH of
11.9435. What was the initial [(CH3)2NH]?
Therefore as the reaction went from its initial state to an equilibrium state, the
concentration of OH minus increased by 0.00878 molar.
(CH3 )2 NH(aq)  H2O( l )
[I]
[C]
[E]
x


(CH3 ) 2 NH2(aq)
 OH(aq)
0
0
+0.00878
0.00878
The weak base dimethylamine, (CH3)2NH, has a Kb value of
5.1 × 10–4. A particular solution of dimethlyamine has a pH of
11.9435. What was the initial [(CH3)2NH]?
We can use this, along with the coefficients in the equation, to find the changes in
the concentrations of the other two species.
Moved to
(CH3 )2 NH(aq)  H2O( l )
[I]
[C]
[E]
x
the
Right


(CH3 ) 2 NH2(aq)
 OH(aq)
0
0
+0.00878
0.00878
The weak base dimethylamine, (CH3)2NH, has a Kb value of
5.1 × 10–4. A particular solution of dimethlyamine has a pH of
11.9435. What was the initial [(CH3)2NH]?
Because the concentration of the product, OH minus increased, this means the
reaction moved to the right as it approached equilibrium.
Moved to
(CH3 )2 NH(aq)  H2O( l )
[I]
[C]
[E]
the
Right


(CH3 ) 2 NH2(aq)
 OH(aq)
x
0
0
–0.00878
+0.00878
+0.00878
0.00878
The weak base dimethylamine, (CH3)2NH, has a Kb value of
5.1 × 10–4. A particular solution of dimethlyamine has a pH of
11.9435. What was the initial [(CH3)2NH]?
Therefore the concentration of the other product, (CH3)2NH2+ , also increased
Moved to
(CH3 )2 NH(aq)  H2O( l )
[I]
[C]
[E]
the
Right


(CH3 ) 2 NH2(aq)
 OH(aq)
x
0
0
–0.00878
+0.00878
+0.00878
0.00878
The weak base dimethylamine, (CH3)2NH, has a Kb value of
5.1 × 10–4. A particular solution of dimethlyamine has a pH of
11.9435. What was the initial [(CH3)2NH]?
And the concentration of the reactant, (CH3)2NH, decreased.
(CH3 )2 NH(aq)  H2O( l )
[I]
[C]
[E]
x
–0.00878


(CH3 ) 2 NH2(aq)
 OH(aq)
1
0
+0.00878
1
+0.00878
0.00878
The weak base dimethylamine, (CH3)2NH, has a Kb value of
5.1 × 10–4. A particular solution of dimethlyamine has a pH of
11.9435. What was the initial [(CH3)2NH]?
The coefficients on (CH3)2NH2+ and OH minus are both 1
0
(CH3 )2 NH(aq)  H2O( l )
[I]
[C]
[E]
x
–0.00878


(CH3 ) 2 NH2(aq)
 OH(aq)
1
0
+0.00878
1
0
+0.00878
0.00878
The weak base dimethylamine, (CH3)2NH, has a Kb value of
5.1 × 10–4. A particular solution of dimethlyamine has a pH of
11.9435. What was the initial [(CH3)2NH]?
So we can say that the concentration of (CH3)2NH2+ increased by 0.00878 molar,
just like the concentration of OH minus.
(CH3 )2 NH(aq)  H2O( l )
[I]
[C]
[E]
1
x
–0.00878


(CH3 ) 2 NH2(aq)
 OH(aq)
1
0
+0.00878
1
0
+0.00878
0.00878
The weak base dimethylamine, (CH3)2NH, has a Kb value of
5.1 × 10–4. A particular solution of dimethlyamine has a pH of
11.9435. What was the initial [(CH3)2NH]?
And because the coefficient on the reactant (CH3)2NH is also 1, like the coefficients
on (CH3)2NH2+ , and OH– ,
(CH3 )2 NH(aq)  H2O( l )
[I]
[C]
[E]
1
x
–0.00878


(CH3 ) 2 NH2(aq)
 OH(aq)
1
0
+0.00878
1
0
+0.00878
0.00878
The weak base dimethylamine, (CH3)2NH, has a Kb value of
5.1 × 10–4. A particular solution of dimethlyamine has a pH of
11.9435. What was the initial [(CH3)2NH]?
We can say that the concentration of (CH3)2NH decreased by 0.00878 molar.
(CH3 )2 NH(aq)  H2O( l )
[I]
[C]
[E]


(CH3 ) 2 NH2(aq)
 OH(aq)
x
0
0
–0.00878
+0.00878
+0.00878
0.00878
The weak base dimethylamine, (CH3)2NH, has a Kb value of
5.1 × 10–4. A particular solution of dimethlyamine has a pH of
11.9435. What was the initial [(CH3)2NH]?
So we have now finished filling in the row for changes in concentration.
(CH3 )2 NH(aq)  H2O( l )
[I]
[C]
[E]


(CH3 ) 2 NH2(aq)
 OH(aq)
x
0
0
–0.00878
+0.00878
+0.00878
0.00878
The weak base dimethylamine, (CH3)2NH, has a Kb value of
5.1 × 10–4. A particular solution of dimethlyamine has a pH of
11.9435. What was the initial [(CH3)2NH]?
And we can now finish the row for equilibrium concentrations.
(CH3 )2 NH(aq)  H2O( l )
[I]
[C]
[E]


(CH3 ) 2 NH2(aq)
 OH(aq)
x
0
0
–0.00878
+0.00878
+0.00878
0.00878
The weak base dimethylamine, (CH3)2NH, has a Kb value of
5.1 × 10–4. A particular solution of dimethlyamine has a pH of
11.9435. What was the initial [(CH3)2NH]?
The equilibrium concentration of (CH3)2NH2+ is 0 plus 0.00878
(CH3 )2 NH(aq)  H2O( l )
[I]
[C]
[E]


(CH3 ) 2 NH2(aq)
 OH(aq)
x
0
0
–0.00878
+0.00878
+0.00878
0.00878
0.00878
The weak base dimethylamine, (CH3)2NH, has a Kb value of
5.1 × 10–4. A particular solution of dimethlyamine has a pH of
11.9435. What was the initial [(CH3)2NH]?
Which is 0.00878 molar
(CH3 )2 NH(aq)  H2O( l )
[I]
[C]
[E]


(CH3 ) 2 NH2(aq)
 OH(aq)
x
0
0
–0.00878
+0.00878
+0.00878
0.00878
0.00878
The weak base dimethylamine, (CH3)2NH, has a Kb value of
5.1 × 10–4. A particular solution of dimethlyamine has a pH of
11.9435. What was the initial [(CH3)2NH]?
The equilibrium concentration of (CH3)2NH is x minus 0.00878
(CH3 )2 NH(aq)  H2O( l )
[I]
[C]
[E]


(CH3 ) 2 NH2(aq)
 OH(aq)
x
0
0
–0.00878
+0.00878
+0.00878
x–0.00878
0.00878
0.00878
The weak base dimethylamine, (CH3)2NH, has a Kb value of
5.1 × 10–4. A particular solution of dimethlyamine has a pH of
11.9435. What was the initial [(CH3)2NH]?
Which we’ll write simply as x minus 0.00878 here.
(CH3 )2 NH(aq)  H2O( l )
[I]
[C]
[E]


(CH3 ) 2 NH2(aq)
 OH(aq)
x
0
0
–0.00878
+0.00878
+0.00878
x–0.00878
0.00878
0.00878
The weak base dimethylamine, (CH3)2NH, has a Kb value of
5.1 × 10–4. A particular solution of dimethlyamine has a pH of
11.9435. What was the initial [(CH3)2NH]?
We now have the equilibrium concentrations of all three species. In order to solve
for x, we substitute these into the Kb expression for dimethylamine.
K b(CH3 )2NH
 (CH 3 ) 2NH 2   OH  



 (CH 3 )2NH

 0.00878 
2
 (CH 3 )2NH
2
0.00878 

K b(CH ) NH 
 x  0.00878 
2
0.00878


x

0.00878



3 2
K b(CH3 )2NH
The weak base dimethylamine, (CH3)2NH, has a Kb value of
5.1 × 10–4. A particular solution of dimethlyamine has a pH of
11.9435. What was the initial [(CH3)2NH]?
We use the equilibrium equation to write the Kb expression for the weak base,
(CH3)2NH .It’s the concentration of (CH3)2NH2+ times [OH-] over the [(CH3)2NH]
K b(CH3 )2NH
 (CH 3 ) 2 NH 2   OH  



 (CH 3 )2NH

 0.00878 
2
 (CH 3 )2NH
2
0.00878 

K b(CH ) NH 
 x  0.00878 
2
0.00878


x

0.00878



3 2
K b(CH3 )2NH
The weak base dimethylamine, (CH3)2NH, has a Kb value of
5.1 × 10–4. A particular solution of dimethlyamine has a pH of
11.9435. What was the initial [(CH3)2NH]?
Substituting 0.00878 in for the concentrations of (CH3)2NH2+ and [OH-] we get that
their product is 0.00878 squared.
K b(CH3 )2NH
 (CH 3 ) 2NH 2   OH  



 (CH 3 )2NH

 0.00878 
2
 (CH 3 )2NH
2
0.00878 

K b(CH ) NH 
 x  0.00878 
2
0.00878


x

0.00878



3 2
K b(CH3 )2NH
The weak base dimethylamine, (CH3)2NH, has a Kb value of
5.1 × 10–4. A particular solution of dimethlyamine has a pH of
11.9435. What was the initial [(CH3)2NH]?
And we’ll substitute x – 0.00878 in for the [(CH3)2NH]
K b(CH3 )2NH
 (CH 3 ) 2NH 2   OH  



 (CH 3 )2NH

 0.00878 
2
 (CH 3 )2NH
2
0.00878 

K b(CH ) NH 
 x  0.00878 
2
0.00878


x

0.00878



3 2
K b(CH3 )2NH
The weak base dimethylamine, (CH3)2NH, has a Kb value of
5.1 × 10–4. A particular solution of dimethlyamine has a pH of
11.9435. What was the initial [(CH3)2NH]?
So now we have the expression, the Kb of dimethylamine equals 0.00878 squared
over x minus 0.00878.
K b(CH3 )2NH
 (CH 3 ) 2NH 2   OH  



 (CH 3 )2NH

 0.00878 
2
 (CH 3 )2NH
2
0.00878 

K b(CH ) NH 
 x  0.00878 
2
0.00878


x

0.00878



3 2
K b(CH3 )2NH
The weak base dimethylamine, (CH3)2NH, has a Kb value of
5.1 × 10–4. A particular solution of dimethlyamine has a pH of
11.9435. What was the initial [(CH3)2NH]?
Now, we’ll rearrange this equation to solve for x minus 0.00878. We get x minus
0.00878 equals 0.00878 squared over the Kb of dimethylamine.
 (CH 3 ) 2NH 2   OH  


K b(CH3 )2NH  
2
 (CH 3 )2NH
0.00878


x

0.0087
8



2
K b(CH3 )2NH
0.00878 


2
 (CH 3 )2NH
0.00878 

 0.15115
 x  0.00878  
2
4
5.1  10
0.00878 

K b(CH3 )2NH 
x  0.15115  0.00878
 x  0.00878 
x  0.15993
2
0.00878 

 x  0.00878  
 (CH 3 )2 NHinitial  x  0.16 M
K b(CH3 )2NH
The weak base dimethylamine, (CH3)2NH, has a Kb value of
5.1 × 10–4. A particular solution of dimethlyamine has a pH of
11.9435. What was the initial [(CH3)2NH]?
We’ll re-write this equation up here.
 (CH 3 ) 2NH 2   OH  


K b(CH3 )2NH  
2
 (CH 3 )2NH
0.00878


x

0.00878



2
K b(CH3 )2NH
0.00878 


2
 (CH 3 )2NH
0.00878 

 0.15115
 x  0.00878  
2
4
5.1  10
0.00878 

K b(CH3 )2NH 
x  0.15115  0.00878
 x  0.00878 
x  0.15993
2
0.00878 

 x  0.00878  
 (CH 3 )2 NHinitial  x  0.16 M
K b(CH3 )2NH
The weak base dimethylamine, (CH3)2NH, has a Kb value of
5.1 × 10–4. A particular solution of dimethlyamine has a pH of
11.9435. What was the initial [(CH3)2NH]?
We’ll substitute 5.1 × in for the Kb of dimethylamine
 (CH 3 ) 2NH 2   OH  


K b(CH3 )2NH  
2
 (CH 3 )2NH
0.00878


x

0.00878



2
K b(CH3 )2NH
0.00878 


2
 (CH3 )2NH
0.00878 

 0.15115
 x  0.00878  
2
4
0.00878 
5.1  10

K b(CH3 )2NH 
x  0.15115  0.00878
 x  0.00878 
x  0.15993
2
0.00878 

 x  0.00878  
 (CH 3 )2 NHinitial  x  0.16 M
K b(CH3 )2NH
The weak base dimethylamine, (CH3)2NH, has a Kb value of
5.1 × 10–4. A particular solution of dimethlyamine has a pH of
11.9435. What was the initial [(CH3)2NH]?
0.00878 squared divided by 5.1 × 10-4 is equal to 0.15115. Or we can simplify this
and say that…
 (CH 3 ) 2NH 2   OH  


K b(CH3 )2NH  
2
 (CH 3 )2NH
0.00878


x

0.00878



2
K b(CH3 )2NH
0.00878 


2
 (CH3 )2NH
0.00878 

 0.15115
 x  0.00878  
2
4
0.00878 
5.1  10

K b(CH3 )2NH 
x  0.15115  0.00878
 x  0.00878 
x  0.15993
2
0.00878 

 x  0.00878  
 (CH 3 )2 NHinitial  x  0.16 M
K b(CH3 )2NH
The weak base dimethylamine, (CH3)2NH, has a Kb value of
5.1 × 10–4. A particular solution of dimethlyamine has a pH of
11.9435. What was the initial [(CH3)2NH]?
x minus 0.00878 is equal to 0.15115
 (CH 3 ) 2NH 2   OH  


K b(CH3 )2NH  
2
 (CH 3 )2NH
0.00878


x

0.00878



2
K b(CH3 )2NH
0.00878 


2
 (CH3 )2NH
0.00878 

 x  0.00878  + 0.008784  0.15115
2
+0.00878
0.00878 
5.1  10

K b(CH3 )2NH 
x  0.15115  0.00878
 x  0.00878 
x  0.15993
2
0.00878 

 x  0.00878  
 (CH 3 )2 NHinitial  x  0.16 M
K b(CH3 )2NH
The weak base dimethylamine, (CH3)2NH, has a Kb value of
5.1 × 10–4. A particular solution of dimethlyamine has a pH of
11.9435. What was the initial [(CH3)2NH]?
In order to isolate x, we add 0.00878 to both sides of the equation..
 (CH 3 ) 2NH 2   OH  


K b(CH3 )2NH  
2
 (CH 3 )2NH
0.00878


x

0.00878



2
K b(CH3 )2NH
0.00878 


2
 (CH3 )2NH
0.00878 

 x  0.00878  + 0.008784  0.15115
2
+0.00878
0.00878 
5.1  10

K b(CH3 )2NH 
x  0.15115  0.00878
 x  0.00878 
x  0.15993
2
0.00878 

 x  0.00878  
 (CH 3 )2 NHinitial  x  0.16 M
K b(CH3 )2NH
The weak base dimethylamine, (CH3)2NH, has a Kb value of
5.1 × 10–4. A particular solution of dimethlyamine has a pH of
11.9435. What was the initial [(CH3)2NH]?
The negative 0.00878 and positive 0.00878 on the left side add up to zero,
 (CH 3 ) 2NH 2   OH  


K b(CH3 )2NH  
2
 (CH 3 )2NH
0.0087
8


x

0.00878



2
K b(CH3 )2NH
0.00878 


2
 (CH3 )2NH
0.00878 

 x  0.00878  + 0.008784  0.15115
2
+0.00878
0.00878 
5.1  10

K b(CH3 )2NH 
x  0.15115  0.00878
 x  0.00878 
x  0.15993
2
0.00878 

 x  0.00878  
 (CH 3 )2 NHinitial  x  0.16 M
K b(CH3 )2NH
The weak base dimethylamine, (CH3)2NH, has a Kb value of
5.1 × 10–4. A particular solution of dimethlyamine has a pH of
11.9435. What was the initial [(CH3)2NH]?
So they’ll cancel out, and we left with x equals 0.15115 plus 0.00878, which we’ll
write here.
 (CH 3 ) 2NH 2   OH  


K b(CH3 )2NH  
2
 (CH 3 )2NH
0.00878


x

0.00878



2
K b(CH3 )2NH
0.00878 


2
 (CH3 )2NH
0.00878 

 0.15115
 x  0.00878  
2
4
0.00878 
5.1  10

K b(CH3 )2NH 
x  0.15115  0.00878
 x  0.00878 
x  0.15993
2
0.00878 

 x  0.00878  
 (CH 3 )2 NHinitial  x  0.16 M
K b(CH3 )2NH
The weak base dimethylamine, (CH3)2NH, has a Kb value of
5.1 × 10–4. A particular solution of dimethlyamine has a pH of
11.9435. What was the initial [(CH3)2NH]?
Adding these gives us a total of 0.15993
 (CH 3 ) 2NH 2   OH  


K b(CH3 )2NH  
2
 (CH 3 )2NH
0.00878


x

0.00878



2
K b(CH3 )2NH
0.00878 


2
 (CH3 )2NH
0.00878 

 0.15115
 x  0.00878  
2
4
0.00878 
5.1  10

K b(CH3 )2NH 
x  0.15115  0.00878
 x  0.00878 
x  0.15993  0.16
2
0.00878 

 x  0.00878  
 (CH 3 )2 NHinitial  x  0.16 M
K b(CH3 )2NH
The weak base dimethylamine, (CH3)2NH, has a Kb value of
5.1 × 10–4. A particular solution of dimethlyamine has a pH of
11.9435. What was the initial [(CH3)2NH]?
The given value for Kb has 2 significant figures, so we round 0.15993 to 2
significant figures and we get 0.16
0.00878 

 x  0.00878  
K b(CH3 )2NH
2
 x  0.00878 
0.00878 


2
4
 0.15115
5.1  10
x  0.15115  0.00878
x  0.15993  0.16
 (CH 3 )2 NHinitial  x  0.16 M
The weak base dimethylamine, (CH3)2NH, has a Kb value of
5.1 × 10–4. A particular solution of dimethlyamine has a pH of
11.9435. What was the initial [(CH3)2NH]?
In the ICE table, we had defined x as the initial concentration of dimethlyamine
0.00878 

 x  0.00878  
K b(CH3 )2NH
2
 x  0.00878 
0.00878 


2
4
 0.15115
5.1  10
x  0.15115  0.00878
x  0.15993  0.16
 (CH 3 )2 NHinitial  x  0.16 M
The weak base dimethylamine, (CH3)2NH, has a Kb value of
5.1 × 10–4. A particular solution of dimethlyamine has a pH of
11.9435. What was the initial [(CH3)2NH]?
So we can say that the initial concentration of dimethlyamine equals x, equals 0.16
molar, or more simply, initial concentration of dimethlyamine…
0.00878 

 x  0.00878  
K b(CH3 )2NH
2
 x  0.00878 
0.00878 


2
4
 0.15115
5.1  10
x  0.15115  0.00878
x  0.15993  0.16
 (CH 3 )2 NHinitial  =x  0.16 M
The weak base dimethylamine, (CH3)2NH, has a Kb value of
5.1 × 10–4. A particular solution of dimethlyamine has a pH of
11.9435. What was the initial [(CH3)2NH]?
equals 0.16 molar.
0.00878 

 x  0.00878  
K b(CH3 )2NH
2
 x  0.00878 
0.00878 


2
4
 0.15115
5.1  10
x  0.15115  0.00878
x  0.15993  0.16
 (CH 3 )2 NHinitial  =x  0.16 M
The weak base dimethylamine, (CH3)2NH, has a Kb value of
5.1 × 10–4. A particular solution of dimethlyamine has a pH of
11.9435. What was the initial [(CH3)2NH]?
So we’ve now answered the question this problem was asking. The initial
concentration of (CH3)2NH, or dimethlyamine was 0.16 molar.