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, K_a

  ⚛ Calculate the equilibrium concentration of hydrogen ions, [H⁺_(aq)], by substituting the values for K_a 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, K_b.

  ⚛ Calculate the value of the base dissociation constant, K_b, using K_a and K_w
  (see Hydrolysis of Acids and Bases)

  ⚛ Calculate the equilibrium concentration of hydroxide ions, [OH^-_(aq)], by substituting the values for K_b 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.

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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
  (assumes additivity of volumes)

  ⚛ 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
  (assumes additivity of volumes)

  ⚛ 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
  (assuming additivity of volumes)

  ⚛ 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) + H_2(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
  (assuming additivity of volumes)

  ⚛ 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) (K_b) using the acid dissociation constant for HA (K_a) and the value for the self-dissociation of water (K_w):

  K_b = K_w ÷ K_a
  Note: at 25°C, K_w = 10^-14, so

  K_b = 10^-14 ÷ K_a (at 25°C)

  ⚛ Calculate the equilibrium concentration of hydroxide ions ([OH^-_(eq)] = x) using the calculated values for K_b 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)

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

Solution:

Write the balanced chemical equation for the neutralisation reaction:

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:

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

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:

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:

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, K_a = 1.8 × 10^-5, and the equilibrium concentrations from the R.I.C.E.Table above.

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.

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

Solution

Write the balanced chemical equation for the neutralisation reaction.

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

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:

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.

Determine the value of K_b

Unfortunately you are unlikely to find the value of K_b for acetate ions tabulated.
But you will find the value of the acid dissociation constant for acetic acid tabulated, K_a(CH₃COOH_(aq)) = 1.8 × 10^-5

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

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

K_w = [H⁺_(aq)][OH^-_(aq)]

At 25°C, K_w = 10^-14

So now we can write:

K_a × K_b = 10^-14

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

(1.8 × 10^-5) × K_b = 10^-14

K_b = (10^-14) ÷ (1.8 × 10^-5) = 5.6 × 10^-10

Therefore:

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:

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

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!

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

Solution:

Write the balanced chemical equation for the neutralisation reaction:

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