Working with Acid-Base Equilibria When a weak acid and a weak base are mixed, an equilibrium is...

Working withAcid-Base Equilibria

When a weak acid and a weak base are mixed, an equilibrium is always established.

Example 2Starting With

Amphiprotic Anions

The hydrogen oxalate ion, HC2O4

–, is mixed with the hydrogen sulphite ion,

HSO3–.

We’re told that the hydrogen oxalate ion, HC2O4–, is mixed with the hydrogen

sulphite ion, HSO3–.

The hydrogen oxalate ion, HC2O4

–, is mixed with the hydrogen sulphite ion,

HSO3–.

Complete the equation for the equilibrium that is

established.And we’re asked to complete the equation for the equilibrium that is

established.

The hydrogen oxalate ion, HC2O4–, is mixed with the hydrogen sulphite

ion, HSO3–. Complete the equation for the equilibrium that is established.

We start by writing the formulas for the two given reactants, HC2O4 – and HSO3 –. We find HC2O4 minus on the acid table…

3 222

4 2 4 3H C OC O H SO HSO



And we see that its on both sides of the table so it can act as either an acid or a base.

Weak Acid

Weak Base



Therefore, HC2O4 minus is amphiprotic

Weak Acid

Weak BaseHC2O4

– is Amphiprotic



We’ll label it as amphiprotic here.

3 222

4 2 4 3H C OC O HSO SO H

Amphiprotic



Now we’ll look up HSO3 minus on the table…

3 222

4 2 4 3H C OC O H H S O SO

Amphiprotic



We see that HSO3 minus is also on both sides of the table.

Weak Acid

Weak Base



So it is also amphiprotic

Weak Acid

Weak BaseHSO3– is

Amphiprotic



We’ll label it as amphiprotic here.

3 222


Amphiprotic Amphiprotic



Because HC2O4 minus is amphiprotic, it is able to act either as an acid or as a base.

3 222

4 2 4 3H C OC O H H SO O S


Can act as an acid or a base



And because HSO3 minus is amphiprotic, it is also able to act either as an acid or as a base.

3 222

4 2 4 3H C OC O H H SO OS






So the question is, which one of these will act as the acid in the forward reaction? We find out by locating both of these on the left side of the acid table and comparing their strengths.

3 222


Which one will act as the acid?



Locating them, we see that HC2O4 minus is a stronger acid than HSO3 minus.

Weaker Acid

Stronger Acid



The stronger acid of the two, the HC2O4 minus, is the one that will play the role of the acid.

Weaker Acid

Stronger Acid

HC2O4– is stronger,

so it will act as the acid



So we’ll label HC2O4 minus as the acid on the left side of the equation.

3 222

4 2 4 3H C OC O H H SO O S

WrA WrB SrB

SrA



And the HSO3 minus will have to play the role of the base.

3 222


WrA WrB SrB

SrA



A proton is transferred from the acid to the base.

3 222


WrA WrB SrB

SrA

H+



Because HC2O4 minus loses a proton, it will form C2O4 2- .

3 22

2 4 2 4 3HC O C O H H O SOS

WrA WrB SrB

SrA

H+



And Because HSO3 minus gains a proton, it will form H2SO3 .

3 22

2 4 2 4 3HC O C O HSO H SO

WrA WrB SrB

SrA

H+



So we’ve now answered the first question. This is the equilibrium equation.

22 4 3 2 4 2 3HC O HSO C O H SO

WrA WrB SrB

SrAThe Equilibrium Equation

Will Reactants or Products be Favoured at Equilibrium?

The next question is, “Will reactants or products be favoured at equilibrium?”

22 4 3 2 4 2 3HC O HSO C O H SO

WrA WrB SrB

SrAWill Reactants or Products

be Favoured at Equilibrium?


To answer that, we start by identifying the acid on the left side and the acid on the right side.

22 4 3 2 4 2 3HC O HSO C O H SO

WrA WrB SrB

SrAIdentify the acid on the left

side and the acid on the right side.


We already know that the acid on the left side is HC2O4 minus.

22 4 3 2 4 2 3HC O HSO C O H SO

WrA WrB SrB

SrA


To find the acid on the right, we consider the reverse reaction.

22 4 3 2 4 2 3HC O HSO C O H SO

WrA WrB SrB

SrA

Reverse reactio

n


As the reverse reaction occurs, the H2SO3 is the species that loses a proton.

3 22

2 4 2 4 3HC O C O HSO H O S

WrA WrB SrB

SrA

Reverse reactio

n


So H2SO3 is the Acid on the right side.

3 22

2 4 2 4 3HC O C O HSO H O S

WrA WrB SrB

SrA

Reverse reactio

n


Now we focus on the two conjugate acids in this equilibrium, HC2O4 minus on the left, and H2SO3 on the right. We look them both up on the Left side of the acid table so see which one is stronger.

32 22

4 2 4 3H HSO C OC O H SO

WrA WrB SrB

SrA


We can see that, of these two acids, H2SO3 is the stronger acid and HC2O4 minus is the weaker acid.

Stronger Acid

Weaker Acid


So we can label H2SO3 as the stronger acid an HC2O4 minus as the weaker acid in our equilibrium equation.

32 22


WrA WrB SrB

SrA


Now we recall that equilibrium will always favour the side with the Weaker acid.

32 22


WrA WrB SrB

SrAEquilibrium will always favour the side with the

Weaker Acid.


Because the weaker acid is on the reactant side, we can say that reactants are favoured at equilibrium. We have now answered the second question.

3 222

4 2 4 3H C OC O H H S OSO

WrA WrB SrB

SrA

Reactants are favoured

Now we’ll look at relative amounts of reactants and products present at equilibrium. Because reactants are favoured,(click) there are more reactants than products at equilibrium.

2 4 3HC O H SO

Reactants are favoured

22 4 2 3C O H SO

2 4 3HC O HSO

Or we can say that reactant concentrations are higher than product concentrations at equilibrium.

Reactant concentrations are higher than product

concentrations.

22 4 2 3C O H SO

2 4 3HC O HSO

The last question we’re asked is whether the value of Keq for this reaction is less than 1 or greater than 1.

For this reaction as written, which of the following is true?

Keq < 1Keq > 1

22 4 2 3C O H SO


Keq < 1 or Keq > 1

2 4 3HC O HSO

Remember that Keq is the ratio of the concentrations of the products to the concentrations of reactants at equilibrium.

22 4 2 3C O H SO


Keq < 1 or Keq > 1

eq

Products < 1

ReactantsK

Ratio of [Products] to [Reactants]

eq

Products

ReactantsK

2 4 3HC O HSO

In this particular reaction we had determined that the reactants are favoured.

22 4 2 3C O H SO


Keq < 1 or Keq > 1

eq

Products

ReactantsK

eq

Products < 1

ReactantsK

Reactants are

Favoured

2 4 3HC O HSO

Inserting the products and reactants into the Keq expression,

22 4 2 3C O H SO


Keq < 1 or Keq > 1

eq

Products < 1

ReactantsK

2 4 3HC O HSO

22 4 2 3C O H SO

eq

Products

ReactantsK

2 4 3HC O HSO

And knowing that reactants are favoured, we see that the numerator is smaller than the denominator.

22 4 2 3C O H SO


Keq < 1 or Keq > 1

eq

Products < 1

ReactantsK

2 4 3HC O HSO

22 4 2 3C O H SO

Numerator

Denominator

Reactants are Favoured

2 4 3HC O HSO

So the value of Keq for this reaction is less than 1. So we have now answered the last question.

22 4 2 3C O H SO


Keq < 1 or Keq > 1

eq

Products < 1

ReactantsK

2 4 3HC O HSO

22 4 2 3C O H SO


< 1

2 4 3HC O HSO

Remember when reactants are favoured, the value of Keq is always less than 1.

22 4 2 3C O H SO


Keq < 1 or Keq > 1

eq

Products < 1

ReactantsK


< 1

Working with Acid-Base Equilibria When a weak acid and a weak base are mixed, an equilibrium is...

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