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.


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.



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,




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.




equation.4. Using an ICE table, calculate [H3O+] or [OH–].

Finally we convert hydronium or hydroxide concentration to pH.




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.


(aq) 2(aq)2(aq)NaNO Na NO

Neutral Spectato

r

WeakBase

Na+ is an alkali metal cation, so it is a neutral spectator.


(aq2(aq) 2(aq))NaNO NONa

Neutral Spectato

r

WeakBase

And we can discard it.


(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.


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


We will need to find the value of its Kb


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.


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.


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,


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.


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.


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.


2

11b(NO )

2.17 10K

Because NO2 minus is a weak base, we can write its hydrolysis equilibrium equation.


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.


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.


2

11b(NO )

2.17 10K

2(aq) 2 ( ) 2(aq) (aq)HNO NO H H O Ol

H+

Weak Base

Forming HNO2


2

11b(NO )

2.17 10K

2(aq) 2 ( ) 2(aq) (aq)HNO NO H H O Ol

H+

And OH minus.


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.


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.


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.


2

11b(NO )

2.17 10K

[I][C]

[E]


Now we’ll fill in what we can in the initial concentration row.


2

11b(NO )

2.17 10K

[I] 0.35 0 0[C] –x +x +x

[E] 0.35–x x x


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.


2

11b(NO )

2.17 10K

[I] 0.35 0 0[C] –x +x +x

[E] 0.35–x x x


Before hydrolysis occurs, we can say that the concentrations of HNO2 and OH minus are both zero.


2

11b(NO )

2.17 10K

[I] 0.35 0 0[C] –x +x +x

[E] 0.35–x x x


Now, we’ll fill in the change in concentration row.


2

11b(NO )

2.17 10K

[I] 0.35 0 0[C] –x +x +x

[E] 0.35–x x x


Because the concentrations of the products are zero, in order to compensate…


2

11b(NO )

2.17 10K

[I] 0.35 0 0[C] –x +x +x

[E] 0.35–x x x


The reaction will move to the right.


2

11b(NO )

2.17 10K

[I] 0.35 0 0[C] –x +x +x

[E] 0.35–x x x


Moves to the Right

So the concentrations of HNO2 and OH minus will both increase


2

11b(NO )

2.17 10K

[I] 0.35 0 0[C] –x +x +x

[E] 0.35–x x x


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


2

11b(NO )

2.17 10K

[I] 0.35 0 0[C] –x +x +x

[E] 0.35–x x x


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.


2

11b(NO )

2.17 10K

[I] 0.35 0 0[C] –x +x +x

[E] 0.35–x x x


Moves to the Right

Like the HNO2 and the OH minus, the coefficient on NO2 minus is 1,


2

11b(NO )

2.17 10K

[I] 0.35 0 0[C] –x +x +x

[E] 0.35–x x x


Moves to the Right

So we can say that the concentration of NO2 minus goes down by x.


2

11b(NO )

2.17 10K

[I] 0.35 0 0[C] –x +x +x

[E] 0.35–x x x


Moves to the Right

Now we can fill in the equilibrium concentration row.


2

11b(NO )

2.17 10K

[I] 0.35 0 0[C] –x +x +x

[E] 0.35–x x x


We’ll start with the OH minus. It will be 0 plus x


2

11b(NO )

2.17 10K

[I] 0.35 0 0[C] –x +x +x

[E] 0.35–x x x


Which is equal to x


2

11b(NO )

2.17 10K

[I] 0.35 0 0[C] –x +x +x

[E] 0.35–x x x


Similarly, the equilibrium concentration of HNO2 will be 0 + x


2

11b(NO )

2.17 10K

[I] 0.35 0 0[C] –x +x +x

[E] 0.35–x x x


Which is equal to x


2

11b(NO )

2.17 10K

[I] 0.35 0 0[C] –x +x +x

[E] 0.35–x x x


Now, we’ll look at the NO2 minus. It started out as 0.35 molar


2

11b(NO )

2.17 10K

[I] 0.35 0 0[C] –x +x +x

[E] 0.35–x x x


And it went down by x,


2

11b(NO )

2.17 10K

[I] 0.35 0 0[C] –x +x +x

[E] 0.35–x x x


So its equilibrium concentration is 035 minus x


2

11b(NO )

2.17 10K

[I] 0.35 0 0[C] –x +x +x

[E] 0.35–x x x


So now we have the equilibrium concentrations of all species.


2

11b(NO )

2.17 10K

[I] 0.35 0 0[C] –x +x +x

[E] 0.35–x x x


Notice that the equilibrium concentration of OH minus is x.


2

11b(NO )

2.17 10K

[I] 0.35 0 0[C] –x +x +x

[E] 0.35–x x x


We’ll make a note of that up here.


2

11b(NO )

2.17 10K

[I] 0.35 0 0[C] –x +x +x

[E] 0.35–x x x


OH x

At this point, we need to solve for the value of x.


2

11b(NO )

2.17 10K

[I] 0.35 0 0[C] –x +x +x

[E] 0.35–x x x


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.


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.


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.


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,


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


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,


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.


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.


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


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


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.


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


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


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.


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


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


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


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.


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


Because we now have a value for the concentration of hydroxide, we add the unit M for molarity.


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.


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.


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.


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 .


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


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 .


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.


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

Here we’ll go through the steps necessary to find the pH of a weak base with a given...

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