 # pH of Aqueous Solution After Mixing Weak Acid and Strong Base Chemistry Tutorial

## Key Concepts

In this tutorial you will learn how to calculate the pH of an aqueous solution after a strong base is mixed with a weak acid (at 25°C).

• Write the balanced chemical equation for the reaction between the weak acid and the strong base (neutralisation reaction)
• Use the reaction stoichiometry (mole ratio) to determine which reactant, if any, is in excess.
• If the weak acid is in excess:

⚛ Use the reaction stoichiometry to calculate the initial concentration of the excess weak acid.

⚛ Use the reaction stoichiometry to calculate the initial concentration of the salt formed.

⚛ Write a balanced chemical equation for the dissociation of the weak acid.

⚛ Use a R.I.C.E. Table to determine the equilibrium concentrations of the weak acid and salt.

⚛ Write the expression for the dissociation of the weak acid, Ka

⚛ Calculate the equilibrium concentration of hydrogen ions, [H+(aq)], by substituting the values for Ka and the equilibrium concentrations of weak acid and its conjugate base (from salt concentration) into the expression for the acid dissociation.

⚛ Calculate the pH of the resultant solution.

Note: resultant solution is acidic: [H+(aq)] > [OH-(aq)] and pH < 7 at 25°C

• If the strong base in excess:

⚛ Use the reaction stoichiometry to determine the concentration of hydroxide ions, [OH-(aq)], in the resultant solution.

⚛ Calculate the pOH of the resultant solution.

⚛ Calculate the pH of the resultant solution.

Note: resultant solution is basic (alkaline): [OH-(aq)] > [H+(aq)] and pH > 7 at 25°C

• If neither the strong base nor the weak acid is in excess :

⚛ Use reaction stoichiometry to determine the initial concentration of the salt.

⚛ Write a balanced chemical equation for the hydrolysis of the anion (equivalent to a base dissociation).

⚛ Use a R.I.C.E. Table to determine the equilibrium concentrations of this anion, its conjugate acid and hydroxide ions.

⚛ Write the expression for the base dissociation of this anion, Kb.

⚛ Calculate the value of the base dissociation constant, Kb, using Ka and Kw
(see Hydrolysis of Acids and Bases)

⚛ Calculate the equilibrium concentration of hydroxide ions, [OH-(aq)], by substituting the values for Kb and equilibrium concentration of anion into the expression for the base dissociation of the anion.

⚛ Calculate the pOH of the resultant solution.

⚛ Calculate the pH of the resultant solution.

Note: resultant solution is basic (alkaline): [OH-(aq)] > [H+(aq)] and pH > 7 at 25°C

You can use these pH calculations to determine the points needed to plot a titration curve.

No ads = no money for us = no free stuff for you!

## Detailed Instructions for How to Calculate the pH of a Solution After Strong Base has been added to Weak Acid

Recall that a strong base, such as an aqueous solution of metal hydroxide, MOH(aq), dissociates completely so the species present in solution are M+(aq) and OH-(aq) as shown in the chemical equation below:

MOH(aq) → M+(aq) + OH-(aq)

We can calculate the moles of MOH(aq) available to react using the known concentration (mol L-1) and volume (L) of MOH(aq):

n(MOH(available)) = c(MOH(aq)) × V(MOH(aq))
(see Molarity Calculations)

Recall that a weak acid, HA(aq), does not dissociate completely in water so the species present in solution are undissociated HA(aq) molecules, as well as anions, A-(aq), and hydrogen ions (H+(aq)) as shown in the chemical equation below:

HA(aq) ⇋ A-(aq) + H+(aq)

We can calculate the moles of HA(aq) available to react using the known concentration (mol L-1) and volume (L) of HA(aq):

n(HA(available)) = c(HA(aq)) × V(HA(aq))
(see Molarity Calculations)

Recall that when a base is added to an acid a neutralisation reaction occurs which produces a salt and water:

HA(aq) + MOH(aq → MA(aq) + H2O(l)

Recall that we can use the stoichiometric ratio (mole ratio) of weak acid to strong base in the balanced chemical equation to determine which reactant, if any, is in excess:

HA(aq) : MOH(aq) is 1:1, so:

• Weak acid is in excess if n(HA(available) > n(MOH(available))

n(HA(excess)) = n(HA(available)) − n(HA(reacted))

Note: n(HA(reacted)) = n(MOH(available)) since all the MOH available has reacted

• Strong base is in excess if n(MOH(available) > n(HA(available))

n(MOH(excess)) = n(MOH(available)) − n(MOH(reacted))

Note: n(MOH(reacted)) = n(HA(available)) since all the HA available has reacted

• Neither strong base nor weak acid is in excess if n(HA(available)) = n(MOH(available))

Note: all the available HA has reacted with all the available MOH.
The species in solution are ONLY due to the dissociation of the salt (MA(aq))

When strong base is added to weak acid the resultant solution can be either acidic or basic.
The resultant solution will not be neutral.
Recall that the condition for a solution to be neutral is when [H+(aq)] = [OH-(aq)]
When a strong base is added to a weak acid this condition for a neutral solution cannot be met.

The resultant solution is

• acidic if the weak acid is in excess. (The strong base is the limiting reagent).

n(HA(available)) > n(MOH(available))

[H+(aq)] > [OH-(aq)]

• basic (alkaline) if the strong base is in excess. (The weak acid is the limiting reagent).
• n(MOH(available)) > n(HA(available))

[OH-(aq)] > [H+(aq)]

• basic (alkaline) if neither the strong base nor the weak acid is in excess

n(HA(available)) = n(MOH(available))

But the species in solution is MA(aq) which is the salt of a weak acid and a strong base so the solution is basic.

[OH-(aq)] > [H+(aq)]

How we calculate the pH of the resultant solution after mixing the strong acid and weak base depends on which reactant species, if any, is in excess:

• Weak Acid is in Excess

⚛ Calculate the concentration of excess weak acid in solution after completion of the neutralisation reaction:

HA(aq) + MOH(aq) → MA(aq) + H2O(l)

[HA(excess)] = n(HA(excess)) ÷ V(total)
where V(total) = volume of acid + volume of base

⚛ Use the stoichiometric ratio (mole ratio) of strong base to salt formed in the neutralisation reaction equation to determine the moles of salt formed

HA(aq) + MOH(aq) → MA(aq) + H2O(l)

If the mole ratio of strong base to weak acid to salt is 1:1:1,
then n(MA(produced)) = n(MOH(available)) = n(HA(reacted))

⚛ Calculate the initial concentration of salt in solution:

[MA(initial)] = n(MA(produced)) ÷ V(total)
where V(total) = volume of acid + volume of base

⚛ Write a balanced chemical equation for the dissociation of this weak acid in solution:

HA(aq) ⇋ H+(aq) + A-(aq)

⚛ Use a R.I.C.E. Table for the dissociation of this excess weak acid in solution AFTER the completion of the neutralisation reaction in order to determine the equilibrium concentrations of acid (HA(eq)), conjugate base (A-(eq)), and hydrogen ions (H+(eq)) in the resultant solution

Note: [HA(initial)] = [HA(excess)]
And, [A-(initial)] = [MA(initial)] (soluble salt: MA(aq) → M+(aq) + A-(aq))

Let x = change in concentration of each species as a result of the dissociation of the excess weak acid.
Note: [HA] will decrease by x
And, [A-] and [H+] will both increase by x

R.I.C.E. Table for the Dissociation of Weak Acid
Reaction HA(aq) A-(aq) + H+(aq)
Initial concentration
(mol L-1)
[HA(initial)] =   [A-(intial))] =   0
Change in concentration
(mol L-1)
x   + x   + x
Equilibrium concentration
(mol L-1)
[HA(eq)] ≈   [A-(eq)] ≈   [H+(eq)] = x

Note: assuming very little dissociation of the weak acid then [HA(initial)] − x ≈ [HA(initial)] = [HA(eq)]
AND assuming x is small compared to [A-(initial)] (ie, x << [A-(initial)]) then [A-(eq)] ≈ [A-(initial)]

⚛ Calculate the concentration of hydrogen ions, [H+(aq)]:

Write the expression for the acid dissociation constant:

 Ka = [A-(eq)][H+(eq)] [HA(eq)]

Substitute in the values for the equilibrium concentration of each species from the R.I.C.E. Table above :

 Ka = [A-(eq)] x [HA(eq)]

Rearrange and solve for x (x = [H+(eq)]) :

 [H+(eq)] = x = Ka[HA(eq)] [A-(eq)]

⚛ Calculate the pH of the resultant solution:

pH = −log10[H+(eq)]

• Strong Base is in Excess

⚛ Calculate the concentration of the excess OH-(aq) in solution after completion of the neutralisation reaction:

HA(aq) + MOH(aq) → MA(aq) + H2O(l)

c(OH-(aq)) = n(OH-(excess)) ÷ V(total)
where V(total) = volume of acid + volume of base

⚛ Calculate the pOH of the resultant solution:

pOH = −log10[OH-(aq)]

⚛ Calculate the pH of the resultant solution:

pH = 14 − pOH (assuming 25°C)

• All of the Strong Base has Neutralised all of the Weak Acid (neither is in excess)

⚛ Calculate the concentration of the salt (MA(aq)) using the stoichiometric ratio of acid to salt from the neutralisation equation:

HA(aq) + MOH(aq) → MA(aq) + H2(l)

HA(aq) : MA(aq) is 1:1

n(MA(aq)) = n(HA(available)) = n(HA(reacted))

c(MA(aq)) = n(MA(aq)) ÷ V(total)

where V(total) = volume acid + volume base

⚛ Write the equation for the hydrolysis of the salt's anion (base dissociation of A-(aq)) :

A-(aq) + H2O(l) ⇋ HA(aq) + OH-

Note: the cation, M+(aq), of a strong base does not undergo hydrolysis, but the anion, A-(aq), of a weak acid does undergo hydrolysis.
Therefore, only the anion (A-) of the salt (MA) will undergo hydrolysis.

Also: The conjugate acid of the base (A-(aq)) is HA(aq) (which is the weak acid we added the strong base to!).

⚛ Use a R.I.C.E. Table to determine the equilibrium concentration of A-(eq), HA(eq) and OH-(eq) in the resultant solution:

Note that for the salt MA: MA(aq) → M+(aq) → A-(aq)

So the initial concentration of A- = [A-(initial)] = [MA(aq)] = c(MA(aq))

Let x = change in concentration of each species due to the hydrolysis of A-

Note: [A-] will decrease by x and [HA] and [OH-] will both increase by x

R.I.C.E. Table for the Base Dissociation of Salt's Anion
Reaction A-(aq) + H2O(l) HA(aq) + OH-(aq)
Initial concentration
(mol L-1)
[A-(initial)] =       0   0
Change in concentration
(mol L-1)
x       + x   + x
Equilibrium concentration
(mol L-1)
[A-(eq)] ≈       [HA(eq)] = x   [OH-(eq)] = x

Assuming x << [A-(initial)] then [A-(eq)] ≈ [A-(initial)]

⚛ Write the expression for the base dissociation of A-(aq) (hydrolysis of A-):

 Kb = [HA(aq)][OH-(aq)][A-]

⚛ Calculate the value of the base dissociation constant for A-(aq) (Kb) using the acid dissociation constant for HA (Ka) and the value for the self-dissociation of water (Kw):

Kb = Kw ÷ Ka
Note: at 25°C, Kw = 10-14, so

Kb = 10-14 ÷ Ka (at 25°C)

⚛ Calculate the equilibrium concentration of hydroxide ions ([OH-(eq)] = x) using the calculated values for Kb and [A-(eq)]

 Kb = [x]2     [A-(eq)] [OH-(eq)] = x = √(Kb[A-(eq)])

⚛ Calculate pOH of resultant solution

pOH = −log10[OH-(aq)]

⚛ Calculate pH of resultant solution

pH = 14 − pOH (assume 25°C)

Do you know this?

Play the game now!

## Worked Example: pH of Resultant Solution When the Weak Acid is in Excess (Strong Base is the Limiting Reagent)

Question:

Beaker A contains 25 mL of 0.10 mol L-1 acetic acid (ethanoic acid), CH3COOH(aq).
Ka = 1.8 × 10-5 (at 25°C)

Beaker B contains 10 mL of 0.10 mol L-1 aqueous sodium hydroxide solution, NaOH(aq).

Calculate the pH of the resultant solution when the contents of Beaker B are added to the contents of Beaker A at 25°C.

Solution:

Write the balanced chemical equation for the neutralisation reaction:

 general word equation acid + base → salt + water word equation acetic acid(ethanoic acid) + sodium hydroxide → sodium acetate(sodium ethanoate) + water balanced chemical equation CH3COOH(aq) + NaOH(aq) → CH3COONa(aq) + H2O(l)

Determine which reactant, if any, is in excess.

The moles of acetic acid in Beaker A can be calculated (n = c × V):

n = c × V

n = moles of CH3COOH(aq)

c = concentration of CH3COOH(aq) = [CH3COOH(aq)] = 0.10 mol L-1

V = volume of CH3COOH(aq) in L = 25 mL = 25 mL ÷ 1000 mL/L = 0.025 L

n(CH3COOH(aq)) = 0.10 × 0.025 = 2.5 × 10-3 mol

Moles of NaOH(aq) in Beaker B can be calculated:

n = c × V

n = NaOH(aq)

c = concentration of = [NaOH(aq)] = 0.10 mol L-1

V = volume of NaOH(aq) in L = 10 mL = 10 mL ÷ 1000 mL/L = 0.010 L

n(NaOH(aq)) = 0.10 × 0.010 = 1.0 × 10-3 mol

The balanced chemical equation above tells us that if we want to neutralise 1 mole of acetic acid, CH3COOH(aq), we would need 1 mole of NaOH(aq).
But we have 2.5 × 10-3 mol acetic acid, so we would need 2.5 × 10-3 mol sodium hydroxide to neutralise all the acid.
We only have 1.0 × 10-3 mol NaOH(aq) available, so NaOH(aq) is the limiting reagent, that is, all of the NaOH(aq) added will react.
Acetic acid, CH3COOH(aq), is the reactant in excess, that is, after the reaction with NaOH(aq) is completed there will still be some unreacted CH3COOH(aq) remaining in the beaker.
Of the available 2.5 × 10-3 mol acetic acid only 1.0 × 10-3 mol will react.

Calculate the moles of acetic acid, CH3COOH(aq) that are in excess.

The moles of excess CH3COOH(aq) are the moles of CH3COOH that will remain in the beaker after the reaction with the base is completed.

n(excess)(CH3COOH) = moles available − moles reacted
n(excess)(CH3COOH) = 2.5 × 10-3 − 1.0 × 10-3
n(excess)(CH3COOH) = 1.5 × 10-3 mol

Calculate the moles of sodium acetate, CH3COONa(aq), that have been produced as a result of the neutralisation reaction.

From the balanced chemical equation for the neutralisation reaction we see that for every 1 mole of CH3COOH(aq) consumed, 1 mole of sodium acetate, CH3COONa(aq), will be produced.
So, when 1.0 × 10-3 mol CH3COOH(aq) reacted with NaOH(aq) in the beaker, 1.0 × 10-3 mol of CH3COONa(aq) was produced.
The changes in the amount of each species in the beaker are summarised below:

   moles available moles that react moles left in solution CH3COOH(aq) + NaOH(aq) ⇋ CH3COONa(aq) + H2O(l) 2.5 × 10-3 1.0 × 10-3 0 − 1.0 × 10-3 − 1.0 × 10-3 + 1.0 × 10-3 1.5 × 10-3 0 1.0 × 10-3

Calculate the concentration of the species present in the solution after the neutralisation reaction is completed by assuming additivity of volumes, that is, the final volume of solution is equal to the volume of acetic acid plus the volume of sodium hydroxide:

final volume = volume(acetic acid) + volume(sodium hydroxide)
final volume = 25 mL + 10 mL
final volume = 35 mL
final volume = 35 mL ÷ 1000 mL/L = 0.035 L

   moles in solution (n) final volume (L) concentration: c = n/V (mol L-1) CH3COOH(aq) + NaOH(aq) ⇋ CH3COONa(aq) + H2O(l) 1.5 × 10-3 0 1.0 × 10-3 0.035 0.035 (1.5 × 10-3)/0.035 = 0.043 (1.0 × 10-3)/0.035 = 0.029

To summarise, if we add 10 mL of 0.10 mol L-1 NaOH(aq) to 25 mL 0.10 mol L-1 CH3COOH(aq) the solution immediately after the completion of the neutralisation reaction is made up of 0.043 mol L-1 CH3COOH(aq) and 0.029 mol L-1 CH3COONa(aq)

But what! There's more!
Acetic acid is a weak acid, some of the acetic acid in the beaker will dissociate to produce "extra" acetate ions and hydrogen ions.

CH3COOH(aq) ⇋ CH3COO-(aq) + H+(aq)

So, we expect the equilibrium concentration of acetic acid in solution to be less than 0.043 mol L-1 and the equilibrium concentration of acetate ions to be more than 0.029 mol L-1.

Calculate the equilibrium concentration of weak acid, anions and hydrogen ions in the solution

Let's call the change in concentration of these species x.

We will use a R.I.C.E. table to help us calculate the equilibrium concentrations of the species in solution:

R.I.C.E. Table for the Dissociation of Acetic Acid
Reaction CH3COOH(aq) CH3COO-(aq) + H+(aq)
Initial Concentration
(mol L-1)
0.043   0.029   0
Change in Concentration
(mol L-1)
x   + x   + x
Equilibrium Concentration
(mol L-1)
0.043 − x   0.029 + x   + x

If we assume that x is very small compared to the concentration of acetic acid(1), then [CH3COOH(aq)] = 0.043 − x ≈ 0.043 mol L-1

If we assume that x is very small compared to the concentration of acetate ions(1), then [CH3COO-(aq)] = 0.029 + x ≈ 0.029 mol L-1

Then the R.I.C.E. Table looks like the one below:

R.I.C.E. Table for the Dissociation of Acetic Acid
Reaction CH3COOH(aq) CH3COO-(aq) + H+(aq)
Initial Concentration
(mol L-1)
0.043   0.029   0
Change in Concentration
(mol L-1)
x   + x   + x
Equilibrium Concentration
(mol L-1)
0.043 − x
≈ 0.043
0.029 + x
≈ 0.029
+ x

Calculate the hydrogen ion concentration in the resultant solution at equilibrium using the value of the acid dissociation constant for acetic acid which is 1.8 × 10-5 at 25°C, Ka = 1.8 × 10-5, and the equilibrium concentrations from the R.I.C.E.Table above.

 Ka = [CH3COO-(aq)][H+(aq)][CH3COOH(aq)] 1.8 × 10-5 = [0.029][x][0.043] (1.8 × 10-5) × [0.043] [0.029] = [x] 2.7 × 10-5 mol L-1 = [x]

The concentration of hydrogen ions in the resultant solution is equal to x :

[H+(aq)] = [x] = 2.7 × 10-5 mol L-1

Calculate the pH of the resultant solution:

pH = −log10[H+(aq)]
pH = −log10[2.7 × 10-5]
pH = 4.6

When strong acid is added to excess weak acid, the solution will be acidic, and for aqueous solutions at 25°C the pH of the resultant solution will be less than 7.

Do you understand this?

Take the test now!

## Worked Example: pH of the Resultant Solution When a Weak Acid is Neutralised by a Strong Base

Question:

Beaker A contains 25 mL of 0.10 mol L-1 acetic acid (ethanoic acid), CH3COOH(aq).
Ka = 1.8 × 10-5 (at 25°C)

Beaker B contains 25 mL of 0.10 mol L-1 aqueous sodium hydroxide solution, NaOH(aq).

Calculate the pH of the resultant solution when the contents of Beaker B are added to the contents of Beaker A at 25°C.

Solution

Write the balanced chemical equation for the neutralisation reaction.

 general word equation: acid + base → salt + water word equation: acetic acid(ethanoic acid) + sodium hydroxide → sodium acetate(sodium ethanoate) + water chemical equation: CH3COOH(aq) + NaOH(aq) → CH3COONa(aq) + H2O

Determine which, if any, reactant is in excess.

Calculate the moles of acetic acid in Beaker A.

n = c × V

n = moles of CH3COOH(aq)

c = concentration of CH3COOH(aq) = [CH3COOH(aq)] = 0.10 mol L-1

V = volume of CH3COOH(aq) in L = 25 mL = 25 mL ÷ 1000 mL/L = 0.025 L

n(CH3COOH(aq)) = 0.10 × 0.025 = 2.5 × 10-3 mol

Calculate the moles of sodium hydroxide in Beaker B

n = c × V

n = moles of NaOH(aq)

c = concentration of NaOH(aq) = [NaOH(aq)] = 0.10 mol L-1

V = volume of NaOH(aq) in L = 25 mL = 25 mL ÷ 1000 mL/L = 0.025 L

n(NaOH(aq)) = 0.10 × 0.025 = 2.5 × 10-3 mol

From the balanced chemical equation for the neutralisation reaction, 1 mole of CH3COOH will react with 1 mole NaOH.
Therefore, 2.5 × 10-3 mol NaOH(aq) will completely "neutralise" 2.5 × 10-3 mol CH3COOH(aq).
Neither the NaOH(aq) nor the CH3COOH(aq) is in excess.

But is the final solution really "neutral"? And what is the pH of this final solution?

Our calculations above suggest that the final solution contains no acetic acid, no sodium hydroxide, but it does contain water and a salt, sodium acetate.

Calculate the moles of sodium acetate produced by the neutralisation reaction.

From the balanced chemical equation above we see that if 2.5 × 10-3 mol of CH3COOH(aq) reacts with 2.5 × 10-3 mol of NaOH(aq) then 2.5 × 10-3 mol of CH3COONa(aq) will be produced.

Calculate the concentration of sodium acetate produced by the neutralisation reaction.

[CH3COONa(produced)] = n(CH3COONa(produced)) ÷ V(total)
V(total) = volume of acid + volume of base = 25 mL + 25 mL = 50 mL = 50 mL ÷ 1000 mL/L = 0.050 L

 [CH3COONa(produced)] = n(CH3COONa(produced))iV(total) [CH3COONa(produced)] = 2.5 × 10-30.050 L [CH3COONa(produced)] = 0.050 mol L-1

Note that sodium acetate (CH3COONa(aq)) is soluble in water so the species in solution are sodium ions (Na+(aq)) and acetate ions (CH3COO-(aq)):

CH3COONa(aq) → Na+(aq) + CH3COO-(aq)

From the balanced chemical equation we see that 1 mole of CH3COONa(aq) produces 1 mole of CH3COO-(aq) in the same solution.
Therefore [CH3COONa(aq)] = [CH3COO-(aq)] = 0.050 mol L-1

Write the chemical equation for the hydrolysis of acetate ions in solution (base dissociation of acetate ions).

Sodium ions do not react with water, do not hydrolyse, but acetate ions will react with water, will hydrolyse!
The chemical equation for the hydrolysis of the acetate ions (ethanoate ions) is shown below:

base + acid conjugate
acid
+ conjugate
base
CH3COO-(aq) + H2O(l) CH3COOH(aq) + OH-(aq)
acetate ions
(ethanoate ions)
+ water acetic acid
(ethanoic acid)
+ hydroxide ions

So the final solution after "neutralising" acetic acid with sodium hydroxide will contain some acetic acid (CH3COOH(aq)) and hydroxide ions (OH-(aq)) after all!
But how much of the acetate ions available will actually hydrolyse to produce hydroxide ions?

Write an expression for the base dissociation constant for acetate ions.

 Kb = [CH3COOH(aq)][OH-(aq)][CH3COO-(aq)]

Determine the value of Kb

Unfortunately you are unlikely to find the value of Kb for acetate ions tabulated.
But you will find the value of the acid dissociation constant for acetic acid tabulated, Ka(CH3COOH(aq)) = 1.8 × 10-5

Look what happens when we multiply Ka for acetic acid by Kb, the base dissociation constant for the conjugate base of acetic acid...

 Ka × Kb = [CH3COO-(aq)][H+(aq)][CH3COOH(aq)] × [CH3COOH(aq)][OH-(aq)][CH3COO-(aq)] Ka × Kb = [H+(aq)] × [OH-(aq)]

And you may recognise the product [H+(aq)][OH-(aq)] as the expression for the self-dissociation of water:

Kw = [H+(aq)][OH-(aq)]

At 25°C, Kw = 10-14

So now we can write:

Ka × Kb = 10-14

And we can use that to calculate the base dissociation constant for the actetate ions (using the value of Ka for its conjugate acid, acetic acid):

(1.8 × 10-5) × Kb = 10-14

Kb = (10-14) ÷ (1.8 × 10-5) = 5.6 × 10-10

Therefore:

 Kb = [CH3COOH(aq)][OH-(aq)][CH3COO-(aq)] = 5.6 × 10-10

Use a R.I.C.E. Table to calculate the equilibruim concentrations of acetate ions, acetic acid molecules and hydroxide ions in the resultant solution.

The intial concentration of acetate ions, [CH3COO-aq], is the concentration of acetate ions produced as a result of the completion of the neutralisation reaction:
[CH3COO-(initial)] = 0.050 mol L-1

Initially, the concentration of acetic acid is 0 because all the of the acid was neutralised by all the base, and at that point there would be no excess of hydroxide ions.
[CH3COOH(initial)] = 0 mol L-1
[OH-(initial)] = 0 mol L-1

Let x be the amount of CH3COO- that undergoes hydrolysis, then the concentration of acetic acid will increase by x and the concentration of OH- will increase by x as shown in the R.I.C.E. Table below:

R.I.C.E. Table for the Hydrolysis of Acetate Ions
Reaction CH3COO-(aq) + H2O(l) CH3COOH(aq) + OH-(aq)
Initial Concentration
(mol L-1)
0.050       0   0
Change in Concentration
(mol L-1)
x       + x   + x
Equilibrium Concentration
(mol L-1)
0.050 − x       x   x
assume x << 0.05
Equilibrium Concentration
(mol L-1)
0.050       x   x

Substitute the values for the equilibrium concentration of each species into the expression for Kb in order to determine the equilibrium concentration of hydroxide ions.

 Kb = [CH3COOH(aq)][OH-(aq)][CH3COO-(aq)] = 5.6 × 10-10 [x][x][0.05] = 5.6 × 10-10 [x]2 = 5.6 × 10-10 × [0.05] √[x]2 = √ 2.8 × 10-11 [x] = 5.3 × 10-6

Therefore the concentration of hydrogen ions in solution is 5.3 × 10-6 mol L-1

[OH-] = [x] = 5.3 × 10-6 mol L-1

Calculate the pOH of this aqueous salt solution:

pOH = −log10[OH-] = −log10[5.3 × 10-6] = 5.3

Calculate the pH of this aqueous salt solution at 25°C:

pH = 14 − pOH = 14 - 5.3 = 8.7

Although we often say that a weak acid has been "neutralised" by a strong base, the resulting solution is NOT neutral because the concentration of hydroxide ions will be greater than the concentration of hydrogen ions:

[OH-(aq)] > [H+(aq)]

The resulting "neutralised" solution of weak acid by strong base is actually basic because the hydroxide ions are present in excess!

Can you apply this?

Join AUS-e-TUTE!

Take the exam now!

## Worked Example: pH of Resultant Solution When the Strong Base is in Excess (Weak Acid is the Limiting Reagent)

Question:

Beaker A contains 25 mL of 0.10 mol L-1 acetic acid (ethanoic acid), CH3COOH(aq).
Ka = 1.8 × 10-5 (at 25°C)

Beaker B contains 30 mL of 0.10 mol L-1 aqueous sodium hydroxide solution, NaOH(aq).

Calculate the pH of the resultant solution when the contents of Beaker B are added to the contents of Beaker A at 25°C.

Solution:

Write the balanced chemical equation for the neutralisation reaction:

 general word equation acid + base → salt + water word equation acetic acid(ethanoic acid) + sodium hydroxide → sodium acetate(sodium ethanoate) + water balanced chemical equation CH3COOH(aq) + NaOH(aq) → CH3COONa(aq) + H2O(l)

Determine which reactant, if any, is in excess.

The moles of acetic acid in Beaker A can be calculated (n = c × V):

n = c × V

n = moles of CH3COOH(aq)

c = concentration of CH3COOH(aq) = [CH3COOH(aq)] = 0.10 mol L-1

V = volume of CH3COOH(aq) in L = 25 mL = 25 mL ÷ 1000 mL/L = 0.025 L

n(CH3COOH(aq)) = 0.10 × 0.025 = 2.5 × 10-3 mol

Moles of NaOH(aq) in Beaker B can be calculated:

n = c × V

n = NaOH(aq)

c = concentration of = [NaOH(aq)] = 0.10 mol L-1

V = volume of NaOH(aq) in L = 30 mL = 10 mL ÷ 1000 mL/L = 0.030 L

n(NaOH(aq)) = 0.10 × 0.030 = 3.0 × 10-3 mol

From the balanced chemical equation above, 1 mol CH3COOH(aq) reacts completely with 1 mol NaOH(aq).
Therefore 2.5 × 10-3 mol CH3COOH(aq) reacts completely with 2.5 × 10-3 mol NaOH(aq).
The available moles of NaOH(aq) in Beaker B is 3.0 × 10-3 mol which is greater than the 2.5 × 10-3 which are need to "neutralise" all the acid, therefore the NaOH(aq) is the reactant in excess.

Calculate the concentration of the excess hydroxide ions, [OH-(aq)].

n(NaOH(excess)) = n(NaOH(available)) − n(NaOH(reacted))
n(NaOH(excess)) = 3.0 × 10-3 − 2.5 × 10-3 = 5.0 × 10-4 mol

V(total) = volume of acid + volume of base (assuming additivity of volumes)
V(total) = 25 mL + 30 mL = 55 mL = 55 mL ÷ 1000 mL/L = 0.055 L

[NaOH(excess)] = n(NaOH(excess)) ÷ V(total)
[NaOH(excess)] = 5.0 × 10-4 mol ÷ 0.055 L = 9.1 × 10-3 mol L-1

Note that NaOH(aq) is a strong base that fully dissociates in water so,
[OH-(excess)] = [NaOH(excess)] = 9.1 × 10-3 mol L-1

Calculate the pOH of the resultant solution

pOH = −log10[OH-(aq)] = −log10[9.1 × 10-3] = 2.0

Calculate the pH of the resultant solution

pH = 14 − pOH (at 25°C)
pH = 14 − 2.0 = 12

Footnotes:

(1) If we don't make this assumption then we will have to solve the quadratic equation:

 Ka = [CH3COO-][H+][CH3COOH] 1.8 × 10-5 = [0.029 + x][x][0.043 − x] 1.8 × 10-5 = 0.029x + x2[0.043 − x] (1.8 × 10-5) × [0.043 − x] = 0.029x + x2 (7.74 × 10-7) − (1.8 × 10-5)x = x2 + 0.029x 0 = x2 + 0.029x + (1.8 × 10-5)x − (7.74 × 10-7) 0 = x2 + 0.029018x − (7.74 × 10-7)

 x = -b ± √(b2 − 4ac)2a x = -0.029018 ± √([0.029018]2 − (4 × 1 × −7.74 × 10-7)2 × 1 x = -0.029018 ± √(8.4514 × 10-4)2 × 1 x must be positive so x = -0.029018 + √(8.4514 × 10-4)2 × 1 x = -0.029018 + 0.029072 × 1 x = 2.7 × 10-5

[CH3COOH] = 0.043 − (2.7 × 10-5) = 0.042973 ≈ 0.043 mol L-1
[CH3COO-] = 0.029 + (2.7 × 10-5) = 0.029027 ≈ 0.029 mol L-1