Acid-Base Reactions (College Board AP® Chemistry)

Study Guide

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Written by: Philippa Platt

Reviewed by: Stewart Hird

Strong Acid Reactions with a Base

Strong Acid - Strong Base

When a strong acid and a strong base are mixed, they react quantitatively in a reaction represented by the equation:

H₃O⁺ (aq) + OH^– (aq) → 2H₂O (l)

H⁺ (aq) + OH^– (aq) → H₂O (l)

An example is HCl reacting with KOH
- HCl + KOH → KCl + H₂O
The pH of the resulting solution may be determined from the concentration of excess reagent

Worked Example

What is the pH of a solution after the addition of 15 mL of 0.200 M KOH to 20 mL of 0.200 M HCl

(K_w = 1.00 x 10^-14 at 25 °C)

Answer:

Step 1: Write out the equation and form the ionic equation

HCl (aq) + KOH (aq) → KCl (aq) + H₂O (l)
H⁺ (aq) + OH^– (aq) → H₂O (l)

Step 2: Calculate the moles of each reactant

Moles of H⁺ = 0.015 L x 0.200 M = 0.003 mol
Moles of OH^– = 0.020 L x 0.200 M = 0.004 mol

Step 3: Determine mols of reactant left over

OH^– is in excess
Moles of OH^– after reaction = 0.004 mol - 0.003 mol = 0.001 mol

Step 4: Calculate the pH of the solution

K_w = [H⁺][OH^–]
OR
[H⁺] = $\frac{K_{w}}{[{OH}^{-}]}$
[H⁺] = $\frac{1.00 \times 10^{- 14}}{0.001}$ = 1.00 x 10^-11
pH = -log10[H⁺] = 11.00 (to 2 d.p)

Weak base - Strong acid

When a weak base and a strong acid are mixed, they will react quantitatively in a reaction represented by the equation:

B (aq) + H₃O⁺ (aq) ⇌ HB⁺ (aq) + H₂O (l)

For example if NH₃ and HCl react
- NH₃ (aq) + HCl (aq) → NH₄Cl (aq)
- NH₃ (aq) + H₃O⁺ (aq) $⇌$ NH₄⁺ (aq) + H₂O
If the weak base is in excess, then a buffer solution is formed, and the pH can be determined from the Henderson–Hasselbalch equation equation:

pOH = pK_b + log $\frac{[B]}{[{BH}^{+}]}$

This is because a salt and acid will be in the remaining solution
If the strong acid is in excess, then the pH can be determined from the moles of excess hydronium ion and the total volume of solution

pH = -log10[H₃O⁺]

pH = -log10[H⁺]

If they are equimolar, then the (slightly acidic) pH can be determined from the equilibrium represented by the equation:

HB⁺ (aq) + H₂O (l) ⇌ B (aq) + H₃O⁺ (aq)

Acid and base reactions will favour the side with the weakest acids and bases

NH₄⁺ (aq) + H₂O $⇌$ NH₃ (aq) + H₃O+ (aq)

NH₄⁺ is a weak acid, so will have a K_a value < 1
Therefore the equilibrium will shift to the left hand side

Worked Example

Calculate the pH of a solution containing 25 mL of 0.150 NH₃ and 15 mL 0.150 of HCl.

(pK_b of ammonia at 25 °C = 4.75)

Answer:

Step 1: Write the full and ionic equation:

NH₃ (aq) + HCl (aq) $⇌$ NH₄Cl (aq)
NH₃ (aq) + HCl (aq) $⇌$ NH₄⁺ + Cl^– (aq)

Step 2: Calculate the moles of each reactant

Moles of HCl = 0.015 L x 0.150 M = 2.25 x 10^-3mol
Moles of NH₃ = 0.025 L x 0.150 M = 3.75 x 10^-3 mol

Step 3: Determine which reactant is in excess

	HCl (aq)	NH₃ (aq)	→	NH₄⁺	Cl^– (aq)
mol at start	2.23 x 10^-3	3.75 x 10^-3		0.00	0.00
mol at end	0.00	1.52 x 10^-3		2.23 x 10^-3	2.23 x 10^-3

Step 4: Calculate the pOH using the following equation

pOH = pK_b + log $\frac{[B]}{[{BH}^{+}]}$

pOH = 4.75 + log $\frac{1.52 \times 10^{- 3}}{2.23 \times 10^{- 3}}$
pOH = 4.58 and pH = 9.41

Strong Base Reactions with an Acid

Strong Base - Weak Acid

When a weak acid and a strong base are mixed, they react quantitatively in a reaction represented by the equation:

HA (aq) + OH^– (aq) ⇌ A^– (aq) + H₂O (l)

If the weak acid is in excess, then a buffer solution is formed, and the pH can be determined from the Henderson-Hasselbalch equation

pH = pK_a + log $\frac{[A^{-}]}{[HA]}$

If the strong base is in excess, then the pH can be determined from the moles of excess hydroxide ion and the total volume of solution

[H⁺] = $\frac{K_{w}}{[{OH}^{-}]}$

If they are equimolar, then the (slightly basic) pH can be determined from the equilibrium represented by the equation:

A^– (aq) + H₂O (l) ⇌ HA (aq) + OH^– (aq)

Worked Example

Calculate the pH of a solution when 50 mL of 0.100 NaOH is added to 75 mL of 0.200 ethanoic acid.

(pK_a of ethanoic acid at 25 °C = 4.74)

Answer:

Step 1: Write the equation

NaOH (aq) + CH₃COOH (aq) → CH₃COONa + H₂O
OH^– (aq) + CH₃COOH (aq) $⇌$ CH₃COO^– + H₂O

Step 2: Calculate the moles of acid and salt in the solution

Moles of OH^– = 0.050 L x 0.100 M = 5.00 x 10^-3 mol
Moles of CH₃COOH = 0.075 L x 0.200 M = 0.015 mol

	OH^– (aq)	CH₃COOH (aq)	$⇌$	CH₃COO^– (aq)	H₂O (l)
mol at start	5.00 x 10^-3	0.015		0.00	0.00
mol at end	0.00	0.01		5.00 x 10^-3	5.00 x 10^-3

Step 3: Calculate pH using the following equation:

pH = pK_a + log $\frac{[A^{-}]}{[HA]}$

pH = 4.74 + log $\frac{5.00 \times 10^{- 3}}{0.01}$ = 4.45

Weak Base - Weak Acid

When a weak acid and a weak base are mixed, they will react to an equilibrium state whose reaction may be represented by the equation:

HA (aq) + B (aq) ⇌ A^– (aq) + HB⁺ (aq)

In an acid base reaction, the equilibrium will favour the side with the weaker acid and weaker base
If the equilibrium constant for the reaction is greater than one (K >1) the equilibrium will favour the formation of products
- This means the weaker acid and weaker base are on the product side
If the equilibrium constant for the reaction is less than one (K <1) the equilibrium will favour the formation of reactants
- This means the weaker acid and weaker base are on the reactant side
The reaction between HSO₄^- and CO₃^2- is an example of a weak acid - weak base reaction
The products are weaker acids and bases than the reactants, so the equilibrium will favour the formation of the products

Weak acid - weak base reaction

The reaction favours the right hand side

If the solutions are equimolar, then we need to consider the K_a and K_b values to determine where the final solution will be acidic, neutral or basic
This can be shown by the reaction between HF and NH₃

Reaction between HF and NH₃

The reaction favours the right hand side

As we can see the K_a and K_b values are greater for the reactants than the products
- This means they are stronger acids and bases
- Therefore the equilibrium will favour the products and K will be greater than 1
The K_a value for NH₄⁺ is greater than the K_b value for F^–, therefore it is better at forming H₃O⁺ ions than the F^– ion is at forming OH^– ions
Therefore the solution will be acidic

Last updated: 25 August 2024

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Acid-Base Reactions (College Board AP® Chemistry)

Study Guide

Strong Acid Reactions with a Base

Strong Acid - Strong Base

Worked Example

Weak base - Strong acid

Worked Example

Strong Base Reactions with an Acid

Strong Base - Weak Acid

Worked Example

Weak Base - Weak Acid

Weak acid - weak base reaction

Reaction between HF and NH₃

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Author: Philippa Platt

Author: Stewart Hird

Acid-Base Reactions (College Board AP® Chemistry)

Study Guide

Strong Acid Reactions with a Base

Strong Acid - Strong Base

Weak base - Strong acid

Strong Base Reactions with an Acid

Strong Base - Weak Acid

Weak Base - Weak Acid

Weak acid - weak base reaction

Reaction between HF and NH3

You've read 0 of your 10 free study guides

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Author: Philippa Platt

Author: Stewart Hird

Reaction between HF and NH₃