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AP®ChemistryCollege BoardStudy GuidesUnit 8: Acids & BasesAcids & BasesAcid-Base Reactions

Acid-Base Reactions (College Board AP® Chemistry)

Study Guide

Philippa Platt

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)

OR

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 

(Kw = 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

  • Kw = [H+][OH–]
    OR
    [H+] = fraction numerator K subscript straight w over denominator open square brackets OH to the power of minus close square brackets end fraction

  • [H+] = fraction numerator 1.00 space cross times 10 to the power of negative 14 end exponent over denominator 0.001 end fraction = 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)rightwards harpoon over leftwards harpoon 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 = pKb + log fraction numerator stretchy left square bracket B stretchy right square bracket over denominator stretchy left square bracket BH to the power of plus stretchy right square bracket end fraction

  • 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+] 

OR

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 rightwards harpoon over leftwards harpoonNH3 (aq) + H3O+ (aq)

  • NH4+ is a weak acid, so will have a Ka 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.

(pKb of ammonia at 25 °C = 4.75)

Answer:

Step 1: Write the full and ionic equation:

  • NH3 (aq) + HCl (aq) rightwards harpoon over leftwards harpoon NH4Cl (aq)

  • NH3 (aq) + HCl (aq) rightwards harpoon over leftwards harpoon NH4+ + Cl– (aq)

Step 2: Calculate the moles of each reactant

  • Moles of HCl = 0.015 L x 0.150 M = 2.25 x 10-3 mol

  • 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 = pKb + log fraction numerator stretchy left square bracket B stretchy right square bracket over denominator open square brackets BH to the power of plus close square brackets end fraction

  • pOH = 4.75 + log begin mathsize 14px style fraction numerator 1.52 cross times 10 to the power of negative 3 end exponent over denominator 2.23 space cross times 10 to the power of negative 3 end exponent end fraction end style

  • 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 = pKa + logfraction numerator stretchy left square bracket A to the power of minus stretchy right square bracket over denominator stretchy left square bracket HA stretchy right square bracket end fraction

  • 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+] = fraction numerator bold italic K subscript bold w over denominator stretchy left square bracket OH to the power of minus stretchy right square bracket end fraction

  • 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.

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

Answer:

Step 1: Write the equation

  • NaOH (aq) + CH3COOH (aq) → CH3COONa + H2O

  • OH– (aq) + CH3COOH (aq) rightwards harpoon over leftwards harpoonCH3COO– + 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)

rightwards harpoon over leftwards harpoon 

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 = pKa + logfraction numerator stretchy left square bracket A to the power of minus stretchy right square bracket over denominator stretchy left square bracket HA stretchy right square bracket end fraction

  • pH = 4.74 + log fraction numerator 5.00 space cross times 10 to the power of negative 3 end exponent over denominator 0.01 end fraction = 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 CO32- 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

weak-acids-and-bases-example-1

The reaction favours the right hand side

  • If the solutions are equimolar, then we need to consider the Ka and Kb 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

weak-acids-and-bases-example-2

The reaction favours the right hand side

  • As we can see the Ka and Kb 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 Ka value for NH4+ is greater than the Kb 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

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    Author
    Reviewer
    Philippa Platt

    Author: Philippa Platt

    Expertise: Chemistry

    Philippa has worked as a GCSE and A level chemistry teacher and tutor for over thirteen years. She studied chemistry and sport science at Loughborough University graduating in 2007 having also completed her PGCE in science. Throughout her time as a teacher she was incharge of a boarding house for five years and coached many teams in a variety of sports. When not producing resources with the chemistry team, Philippa enjoys being active outside with her young family and is a very keen gardener

    Stewart Hird

    Author: Stewart Hird

    Expertise: Chemistry Lead

    Stewart has been an enthusiastic GCSE, IGCSE, A Level and IB teacher for more than 30 years in the UK as well as overseas, and has also been an examiner for IB and A Level. As a long-standing Head of Science, Stewart brings a wealth of experience to creating Topic Questions and revision materials for Save My Exams. Stewart specialises in Chemistry, but has also taught Physics and Environmental Systems and Societies.