pH & pKa (College Board AP® Chemistry)

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Philippa Platt

Written by: Philippa Platt

Reviewed by: Stewart Hird

pH & pKa

pH

  • pH is defined as -log[H3O+]  

    • Note that [H+] is acceptable to use for AP chemistry

  • To calculate [H3O+] from pH we can use log–pH

pKa

  • pKa is a measure of the acidity or basicity of a substance in solution

    • pKa = -log(Ka)

  • To calculate Ka from pKa we can use log–pH

The relationship between pH and pKa

  • The pH of a solution can be related to the pKa of an acid using the Henderson-Hasselbalch equation

pH = pKa + log fraction numerator stretchy left square bracket A to the power of minus stretchy right square bracket over denominator stretchy left square bracket H A stretchy right square bracket end fraction

  • [A] is the concentration of the conjugate base

  • [HA] is the concentration of the acid

pH curves, pH and pKa

  • During a titration buffer regions occur when 

    • Strong acid - weak base

    • Strong base - weak acid

  • At this point, the buffer formed will resist changes in pH so the pH rises gradually as shown in the buffer region

  • The half equivalence point is the stage of the titration at which exactly half the amount of weak acid has been neutralised

  • At the half equivalence point, [acid] = [conjugate base] or [base] = [conjugate acid]

  • For example during the titration of NaOH and ethanoic acid:

    • CH3COOH + OH rightwards harpoon over leftwards harpoon CH3COO + H2O

    • CH3COOH is the acid

    • CH3COO is the conjugate base

The difference between pH and pKa

  • pH is a measure of the acidity or basicity of an aqueous solution

    • A pH below 7.00 indicates a substance is acidic 

    • A pH above 7.00 indicates a substance is basic

  • pKa is used to show the strength of an acid

    • A low pKa value indicates a strong acid

    • A high pKa value indicates a weak acid 

Worked Example

Calculate the pKa of a 0.020 M solution of a weak acid with a pH value of 4.80.

Answer:

Step 1: Write the expression for Ka

  • Kafraction numerator open square brackets straight H to the power of plus close square brackets open square brackets straight A to the power of minus close square brackets over denominator open square brackets HA close square brackets end fraction

Step 2: Calculate [H3O+] and make an ice chart to calculate [HA]

  • [H3O+] = 10-pH = 10-4.80 = 1.58 x 10-5

 

HA

H2O (l)

rightwards harpoon over leftwards harpoon

A (aq)  

+   

H3O+

Initial Conc.

0.020

 

-

 

0.0

 

0.0

Change

0.020 - 1.58 x 10-5

 

-

 

+ 1.58 x 10-5

 

+ 1.58 x 10-5

Equilibrium 

0.0199842

 

-

 

1.58 x 10-5

 

1.58 x 10-5

Step 3: Calculate Ka

  • Ka = fraction numerator 1.58 cross times 10 to the power of negative 5 end exponent space cross times 1.58 cross times 10 to the power of negative 5 end exponent over denominator 0.0199842 end fraction = 1.25 x 10-8

Step 4: Calculate pKa 

  • pKa = -log(1.25 x 10–8)

  • pKa = 7.90

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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.