Working with
Acid-Base Equilibria
Example 3
Starting With Salts
Here we’ll examine the acid-base equilibrium that results when we mix two salts
that have amphiprotic anions.
The salt potassium hydrogen sulphite,
KHSO3, is mixed with the salt sodium
monohydrogen phosphate, Na2HPO4.
We’re given that the salt potassium hydrogen sulphite, KHSO3, is mixed with the
salt sodium monohydrogen phosphate, Na2HPO4.
The salt potassium hydrogen sulphite,
KHSO3, is mixed with the salt sodium
monohydrogen phosphate, Na2HPO4.
Complete the equation for
the equilibrium that is
established.
And we’re asked to complete the equation for the equilibrium that is established.
KHSO3  K


HSO3

Na 2HPO4  2Na


2
HPO4
The salt potassium hydrogen sulphite, KHSO3, is mixed with the
salt sodium monohydrogen phosphate, Na2HPO4. Complete the
equation for the equilibrium that is established.
We begin by taking the salt KHSO3 and writing the dissociation equation for it.
KHSO3  K


HSO3

Na 2HPO4  2Na


2
HPO4
The salt potassium hydrogen sulphite, KHSO3, is mixed with the
salt sodium monohydrogen phosphate, Na2HPO4. Complete the
equation for the equilibrium that is established.
It dissociates into K+
KHSO3  K


HSO3

Na 2HPO4  2Na


2
HPO4
The salt potassium hydrogen sulphite, KHSO3, is mixed with the
salt sodium monohydrogen phosphate, Na2HPO4. Complete the
equation for the equilibrium that is established.
And HSO3 minus.
KHSO3  K

A
spectator
ion

HSO3

Na 2HPO4  2Na


2
HPO4
The salt potassium hydrogen sulphite, KHSO3, is mixed with the
salt sodium monohydrogen phosphate, Na2HPO4. Complete the
equation for the equilibrium that is established.
K+, an alkali metal cation, is a spectator ion.
KHSO3  K

A
spectator
ion

HSO3

Na 2HPO4  2Na


2
HPO4
The salt potassium hydrogen sulphite, KHSO3, is mixed with the
salt sodium monohydrogen phosphate, Na2HPO4. Complete the
equation for the equilibrium that is established.
So we eliminate it.
KHSO3  K


HSO3

Na 2HPO4  2Na


2
HPO4
The salt potassium hydrogen sulphite, KHSO3, is mixed with the
salt sodium monohydrogen phosphate, Na2HPO4. Complete the
equation for the equilibrium that is established.
And all we’re left with the HSO3 minus.
KHSO3  K


HSO3

Na 2HPO4  2Na


2
HPO4
The salt potassium hydrogen sulphite, KHSO3, is mixed with the
salt sodium monohydrogen phosphate, Na2HPO4. Complete the
equation for the equilibrium that is established.
Now we take the other salt, Na2HPO4 and write its dissociation equation.
KHSO3  K


HSO3

Na 2HPO4  2Na


2
HPO4
The salt potassium hydrogen sulphite, KHSO3, is mixed with the
salt sodium monohydrogen phosphate, Na2HPO4. Complete the
equation for the equilibrium that is established.
It dissociates into 2 Na+
KHSO3  K


HSO3

Na 2HPO4  2Na


2
HPO4
The salt potassium hydrogen sulphite, KHSO3, is mixed with the
salt sodium monohydrogen phosphate, Na2HPO4. Complete the
equation for the equilibrium that is established.
And HPO4 2-, the monohydrogen phosphate ion.
KHSO3  K


HSO3

Na 2HPO4  2Na


2
HPO4
A spectator
ion
The salt potassium hydrogen sulphite, KHSO3, is mixed with the
salt sodium monohydrogen phosphate, Na2HPO4. Complete the
equation for the equilibrium that is established.
The Na+ ion, another alkali metal cation, is a spectator ion.
KHSO3  K


HSO3

Na 2HPO4  2Na


2
HPO4
A spectator
ion
The salt potassium hydrogen sulphite, KHSO3, is mixed with the
salt sodium monohydrogen phosphate, Na2HPO4. Complete the
equation for the equilibrium that is established.
So we eliminate that.
KHSO3  K


HSO3

Na 2HPO4  2Na


2
HPO4
The salt potassium hydrogen sulphite, KHSO3, is mixed with the
salt sodium monohydrogen phosphate, Na2HPO4. Complete the
equation for the equilibrium that is established.
Leaving us with only the monohydrogen phosphate ion.

HSO3
2
HPO4
The salt potassium hydrogen sulphite, KHSO3, is mixed with the
salt sodium monohydrogen phosphate, Na2HPO4. Complete the
equation for the equilibrium that is established.
So we’ll move these two ions (click) so they become reactants…

HSO3
2
 HPO4

H 2PO4
2
 SO3
The salt potassium hydrogen sulphite, KHSO3, is mixed with the
salt sodium monohydrogen phosphate, Na2HPO4. Complete the
equation for the equilibrium that is established.
in a new equilibrium equation.

HSO3
2
 HPO4

H 2PO4
2
 SO3
The salt potassium hydrogen sulphite, KHSO3, is mixed with the
salt sodium monohydrogen phosphate, Na2HPO4. Complete the
equation for the equilibrium that is established.
We’ll look up HSO3 minus on the acid table.
Weak Base
Weak Acid
The salt potassium hydrogen sulphite, KHSO3, is mixed with the
salt sodium monohydrogen phosphate, Na2HPO4. Complete the
equation for the equilibrium that is established.
We see that it is on both sides of the table,
Weak Base
HSO3– is
Amphiprotic
Weak Acid
The salt potassium hydrogen sulphite, KHSO3, is mixed with the
salt sodium monohydrogen phosphate, Na2HPO4. Complete the
equation for the equilibrium that is established.
So HSO3 minus is amphiprotic. It can act as an acid or as a base.

HSO3
2
 HPO4

H 2PO4
2
 SO3
Amphiprotic
The salt potassium hydrogen sulphite, KHSO3, is mixed with the
salt sodium monohydrogen phosphate, Na2HPO4. Complete the
equation for the equilibrium that is established.
So we’ll jot that down here.

HSO3
2
 HPO4

H 2PO4
2
 SO3
Amphiprotic
The salt potassium hydrogen sulphite, KHSO3, is mixed with the
salt sodium monohydrogen phosphate, Na2HPO4. Complete the
equation for the equilibrium that is established.
Now, we’ll find HPO4 2- on the table
Weak Base
Weak Acid
The salt potassium hydrogen sulphite, KHSO3, is mixed with the
salt sodium monohydrogen phosphate, Na2HPO4. Complete the
equation for the equilibrium that is established.
We see that it is also on both sides of the table,
HPO42– is
Amphiprotic
Weak Base
Weak Acid
The salt potassium hydrogen sulphite, KHSO3, is mixed with the
salt sodium monohydrogen phosphate, Na2HPO4. Complete the
equation for the equilibrium that is established.
So we can say that HPO4 2- is amphiprotic

HSO3
Amphiprotic
2
 HPO4

H 2PO4
2
 SO3
Amphiprotic
The salt potassium hydrogen sulphite, KHSO3, is mixed with the
salt sodium monohydrogen phosphate, Na2HPO4. Complete the
equation for the equilibrium that is established.
Which we’ll make a note of here.

HSO3
2
 HPO4

H 2PO4
2
 SO3
Which one will
act as the acid?
The salt potassium hydrogen sulphite, KHSO3, is mixed with the
salt sodium monohydrogen phosphate, Na2HPO4. Complete the
equation for the equilibrium that is established.
Because these are both amphiprotic, we now have to determine which one will play
the role of the acid. To do this we compare their strengths on the acid table.
Stronger Acid
Weaker Acid
The salt potassium hydrogen sulphite, KHSO3, is mixed with the
salt sodium monohydrogen phosphate, Na2HPO4. Complete the
equation for the equilibrium that is established.
We see that HSO3 minus is above HPO4 2- on the left side of the table,
Stronger Acid
HSO3– is a stronger
acid than HPO42–
Weaker Acid
The salt potassium hydrogen sulphite, KHSO3, is mixed with the
salt sodium monohydrogen phosphate, Na2HPO4. Complete the
equation for the equilibrium that is established.
so HSO3– is a stronger acid than HPO42–
HSO3– is a stronger
acid than HPO42–

HSO3
WrA
2
 HPO4
WrB

H 2PO4
SrB
2
 SO3
The salt potassium hydrogen sulphite, KHSO3, is mixed with the
salt sodium monohydrogen phosphate, Na2HPO4. Complete the
equation for the equilibrium that is established.
Therefore, HSO3 minus will act as the acid on the left side.
SrA

HSO3
WrA
2
 HPO4
WrB

H 2PO4
SrB
2
 SO3
The salt potassium hydrogen sulphite, KHSO3, is mixed with the
salt sodium monohydrogen phosphate, Na2HPO4. Complete the
equation for the equilibrium that is established.
And the HPO4 2- will have to act as the base on the left side.
SrA
H+

HSO3
WrA
2
 HPO4
WrB

H 2PO4
SrB
2
 SO3
The salt potassium hydrogen sulphite, KHSO3, is mixed with the
salt sodium monohydrogen phosphate, Na2HPO4. Complete the
equation for the equilibrium that is established.
The acid will donate a proton to the base.
SrA
H+

HSO3
WrA
2
 HPO4
WrB

H 2PO4
SrB
2
 SO3
The salt potassium hydrogen sulphite, KHSO3, is mixed with the
salt sodium monohydrogen phosphate, Na2HPO4. Complete the
equation for the equilibrium that is established.
When HSO3– loses a proton it forms SO3 2-, the sulphite ion.
SrA
H+

HSO3
WrA
2
 HPO4
WrB

H 2PO4
SrB
2
 SO3
SrA
The salt potassium hydrogen sulphite, KHSO3, is mixed with the
salt sodium monohydrogen phosphate, Na2HPO4. Complete the
equation for the equilibrium that is established.
And When HPO4 2– gains a proton it forms H2PO4 –, the dihydrogen phosphate ion.

HSO3
WrA
2
 HPO4
WrB

H 2PO4
SrB
2
 SO3
SrA
The Equilibrium Equation
The salt potassium hydrogen sulphite, KHSO3, is mixed with the
salt sodium monohydrogen phosphate, Na2HPO4. Complete the
equation for the equilibrium that is established.
So we’ve now answered the first question. This is the equilibrium equation for this
reaction.

HSO3
2
 HPO4

H 2PO4
2
 SO3
SrB
WrA
WrB
SrA
Are the Reactants or Products
favoured at equilibrium?
Are the Reactants or Products favoured at
equilibrium?
The next question we’re asked is whether the reactants or products are favoured at
equilibrium.

HSO3
WrA
2
 HPO4
WrB

H 2PO4
SrB
2
 SO3
SrA
Are the Reactants or Products favoured at
equilibrium?
To answer this we start by identifying the acids on the left side and the right side.
Remember, the acid on the left side is HSO3 minus.

HSO3
WrA
2
 HPO4
WrB
Reverse
reaction

H 2PO4
SrB
2
 SO3
Are the Reactants or Products favoured at
equilibrium?
To find the acid on the right, we consider the reverse reaction.
SrA

HSO3
WrA
2
 HPO4
WrB
Reverse
reaction

H 2PO4
SrB
2
 SO3
SrA
Are the Reactants or Products favoured at
equilibrium?
In the reverse reaction the H2PO4 minus loses 1 proton as it forms HPO4 2- .

HSO3
WrA
2
 HPO4
WrB
Reverse
reaction

H 2PO4
SrA
Are the Reactants or Products favoured at
equilibrium?
Therefore, H2PO4 minus is the acid on the right side.
2
 SO3
SrA

HSO3
WrA
2
 HPO4
WrB

H 2PO4
SrA
2
 SO3
SrA
Which acid is stronger
and which one is weaker?
Are the Reactants or Products favoured at
equilibrium?
Now we check these two acids on the left side of the acid table to determine which
one is stronger and which one is weaker.
Stronger Acid
Weaker Acid
Are the Reactants or Products favoured at
equilibrium?
We see that HSO3 minus is above H2PO4 minus on the left side of the table,
Stronger Acid
Sr
Wr
Weaker Acid
Are the Reactants or Products favoured at
equilibrium?
so HSO3 minus is the stronger acid H2PO4 minus is the weaker acid.

HSO3
SrA
2
 HPO4
WrB

H 2PO4
WrA
Are the Reactants or Products favoured at
equilibrium?
Which we’ll make a note of here in the equation.
2
 SO3
SrA

HSO3
SrA
2
 HPO4
WrB

H 2PO4
WrA
2
 SO3
SrA
Equilibrium will always favour
the side with the Weaker Acid.
Are the Reactants or Products favoured at
equilibrium?
Remember that equilibrium will always favour the side with the weaker acid.
Equilibrium favours
the Products

HSO3
SrA
2
 HPO4
WrB

H 2PO4
WrA
2
 SO3
SrA
Equilibrium will always favour
the side with the Weaker Acid.
Are the Reactants or Products favoured at
equilibrium?
And in this reaction, the products have the weaker acid, so equilibrium favours the
products in this reaction.
Equilibrium favours
the Products

HSO3
SrA
2
 HPO4
WrB

H 2PO4
WrA
2
 SO3
SrA
Are the Reactants or Products favoured at
equilibrium?
We’ve now answered the second question. Equilibrium favours the products in this
case.

HSO3
2
 HPO4

H 2PO4
2
 SO3
WrAwhichSrA
SrAFor this reaction
WrB as written,
of the following is true?
Keq < 1
Keq > 1
For this reaction as written, which of the
following is true? Keq < 1
Keq > 1
The third question we’re asked is whether the value of Keq for this reaction is less
than 1 or greater than 1.

HSO3
SrA
2
 HPO4
WrB

H 2PO4
WrA
2
 SO3
SrA
Equilibrium favours
the Products
For this reaction as written, which of the
following is true? Keq < 1
Keq > 1
We have determined that equilibrium favours the products in this reaction.

HSO3
SrA
2
 HPO4
WrB

H 2PO4
WrA
2
 SO3
SrA
Equilibrium favours
the Products
For this reaction as written, which of the
following is true? Keq < 1
Keq > 1
Which means we have a (click) greater amount of products than reactants at
equilibrium.

HSO3
SrA
2
 HPO4
WrB
Productss
Product

K eq 
Reactants
tants
 Reac

H 2PO4
WrA
<1
For this reaction as written, which of the
following is true? Keq < 1
Keq > 1
So Keq, which is the ratio of products to reactants,
2
 SO3
SrA

HSO3
2
 HPO4
SrA
WrB

H 2PO4
2WrA

H



PO
SO
2
4 
3 
Products

K eq 
<1
 HSO   HPO
 Reactants



3
2

4 
For this reaction as written, which of the
following is true? Keq < 1
Keq > 1
Is given by this expression.
2
 SO3
SrA
Equilibrium favours
the Products

HSO3
SrA
2
 HPO4
WrB

H 2PO4
2WrA

H



PO
SO
2
4 
3 
Products

K eq 
<1
 HSO   HPO
 Reactants



3
2

4 
For this reaction as written, which of the
following is true? Keq < 1
Keq > 1
And because equilibrium favours the products.
2
 SO3
SrA
Equilibrium favours
the Products

HSO3
SrA
2
 HPO4
WrB

H 2PO4
2WrA

H



PO
SO
2
4 
3 
Products

K eq 
<1
 HSO   HPO
 Reactants



3
2

4 
For this reaction as written, which of the
following is true? Keq < 1
Keq > 1
The Numerator is larger than the denominator
2
 SO3
SrA
Numerator is
larger than
denominator

HSO3
SrA
2
 HPO4
WrB

H 2PO4
2WrA

2
 SO3
SrA
H



PO
SO
2
4 
3 
Products

K eq 
<1 > 1
 HSO   HPO
 Reactants



3
2

4 
For this reaction as written, which of the
following is true? Keq < 1
Keq > 1
So the value of Keq for this reaction is greater than 1.

HSO3
SrA
2
 HPO4
WrB
Products

K eq >
 1
 Reactants

H 2PO4
WrA
<1
For this reaction as written, which of the
following is true? Keq < 1
Keq > 1
And now we’ve answered the third question. Keq > 1
2
 SO3
SrA