Chapter 9: Unit 7. pH

pH

The pH is defined as a formula to describe the acidity and the basicity of a solution. The pH formula is given below:

pH=Log 10 (1/[H3O +) =–Log 10 ([H3O +)

Abbreviated as pH = –Log[H3O + ]

pOH=Log10(1/[OH–]) =–Log 10 ([OH–])

Abbreviated as pOH = –Log[OH–]

Since: Kw= {[ H3O + (aq)] x [OH–(aq)]} = 1.0 x 10–14

pkw= –Log [1.0 x 10–14 ] = 14 = p[H3O + ] + p[OH–]

The pH scale is given below:

pH = 0 to 6 (acidic),

pH = 7 (neutral)

pH = 8 to 14 (basic)

The figure below depicts the scale of the pH with some examples:

https://fineartamerica.com/featured/ph–scale–spencer–sutton.html?product=poster

Other examples of the pH of common materials:

http://www1.lsbu.ac.uk/water/water_dissociation.html

The pH Calculations of Solutions

The pH calculations is illustrated in a You Tube video below:

The pH and the pOH calculations with known molarities of H3O + and OH– examples are given below with some notes:

http://www.everettcc.edu/files/programs/academic–resources/transitional–studies/support/tutoring–center/chemistry/w335–ph–worksheet–3.pdf

Rule of pH significant figures:

Thesignificantfiguresintheconcentrationof[H + ]or[OH–]isequaltothenumberof decimalplacesinthepHorpOHandviceversa

• 1) What is the pH of a 0.0235 M HCl solution?

pH = –log[H + ] = –log(0.0235) = 1.629

Note: the molarity of HCl has 3 significant figures. Therefore, pH value will have 3 digits after the decimal.The number before the decimal is not considered.

• 2) What is the pOH of a 0.0235 M HCl solution?

pH = –log[H + ] = –log(0.0235) = 1.629

pOH = 14.000 – pH = 14.000 – 1.629 = 12.371

Note: the molarity of HCl has 3 significant figures. Therefore, pH value will have 3 digits after the decimal.The number before the decimal is not considered.

• 3) What is the pH of a 6.50 x 10–3 M KOH solution? pOH = –log[OH–] = –log(6.50 x 10–3 ) = 2.187 pH = 14.000 – pOH = 14.000 – 2.187 = 11.813

Note: the molarity of KOHhas 3 significant figures. Therefore, pH value will have 3 digits after the decimal. The number before the decimalis not considered.

• 4) A solution is created by measuring 3.60 x 10–3 moles of NaOH and 5.95 x 10–4 moles of HCl into a container and then water is added until the final volume is 1.00 L. What is the pH of this solution?

Since there is both acid and base we will assume a 1 mole acid:1 mole base ratio of neutralization. There is more base than acid so the leftover base is what will affect the pH of the solution.

3.60 x 10–3 moles – 5.95 x 10–4 moles = 3.01 x 10–3 moles NaOH

3.01 x 10–3 moles NaOH = 3.01 x 10–3 M NaOH

1.00 L soln

pOH = –log[OH–] = –log(3.01 x 10–3 ) = 2.521 pH = 14.000 – pOH = 14.000 – 2.521 = 11.479

Note: the molarities have3 significant figures. Therefore, pH value will have 3 digits after the decimal.The number before the decimal is not considered.

• 5) What is the pH of a 6.2 x 10–5 M NaOH solution?

pOH = –log[OH–] = –log(6.2 x 10–5 ) = 4.21 pH = 14.00 – pOH = 14.00 – 4.21 = 9.79

Note: the molarity of NaOH has 2significant figures.Therefore, pH value will have 2digits after the decimal.The number before the decimal is not considered.

• 6) A solution with a H + concentration of 1.00 x 10–7 M is said to be neutral. Why?

pH = –log[H + ] = –log(1.00 x 10–7 ) = 7.000

pOH = 14.000 – pH = 14.000 – 7.000 = 7.000

Note: the molarity of solutionhas 3 significant figures. Therefore, pH value will have 3 digits after the decimal.The number before the decimal is not considered.

pOH = –log[OH–] = 7.000 we can use this to find the OH– concentration

–log[OH–] = 7.000

[OH–] = 10 – pOH = 10 – 7.000

[OH–] = 1.00 x 10–7 M

The concentrations of H +and OH–are equal, as are the pH and pOH, so the solution must be neutral.

Calculations of H3O + and OH– molarities from known pH and the pOH examples are given below with some notes:

http://sciencelas.weebly.com/uploads/5/7/8/7/57876455/11–11ab_ph_calculations_wkst–key.pdf

• 1. a) What is the hydrogen ion concentration of an aqueous HCl solution that has a pH of 3.0?

pH = 3.0

[H3O + ] = 10 – pH = 10 – 3.0 = 1 x 10 – 3

Note: the pH has 1 digit after decimal.Therefore, [H 3 O + ]molarity value will have 1 significant figure.

• b) What is the hydroxide ion concentration of this same solution?

[H3O + ] = 1 x 10 – 3

[OH–] = K w / [[H 3 O + ] = [1.0 x 10 – 14 ] / [1 x 10 – 3 ] = 1 x 10 –11

Note: the pH has 1 digit after decimal.Therefore, [OH–] molarity value will have 1 significant figure.

• c) Which ion, H + or OH–, is in greater concentration?

[H3O + ] = 1 x 10 – 3

[OH–] = 1 x 10 –11

[H3O + ] is greater than [OH–]

• d) Is this solution acidic or basic?

[H3O + ] is greater than [OH–] and the solution is acidic.

• 2. Find the [H + ] and the [OH–] of a solution with a pH of 3.494

pH of 3.494

[H3O + ] = 10 – pH = 10 – 3.494 = 3.21 x 10 – 4

[OH–] = K w / [[H3O + ] = [1.0 x 10 – 14 ] / [3.206 x 10 – 4 ] = 3.12x10 –11

Note: the pH has 3 digits after decimal.Therefore, [H 3 O + ]and [OH–] molarities valueswill have 3 significant figures.

A Phet simulation for the pH determination is given below:

https://phet.colorado.edu/en/simulation/ph-scalehttps://phet.colorado.edu/sims/html/ph-scale/latest/ph-scale_en.html

Simulation Activity

Directions: Use specific examples to demonstrate each of the following learning goals.

• 1. Determine if a solution is acidic or basic using
  • a. pH
  • b. H3O + /OH–ratio (molecular size representation of just the ions in the water equilibrium)
  • c. Hydronium/Hydroxide concentration
• 2. Relate liquid color to pH.
• 3. Predict if dilution and volume will increase, decrease or not change the pH
• 4. Organize a list of liquids in terms of acid or base strength in relative order with supporting evidence.
• 5. Write the water equilibrium expression. Describe how the water equilibrium varies with pH.

Directions: Use specific examples to demonstrate each of the following learning goals.

• 1. Determine if a solution is acidic or basic using
  • a. pH

• b. molecular representation
• c. Hydronium/Hydroxide concentration
• 2. Relate liquid color to pH.
• 3. Predict if dilution and volume will increase, decrease or not change the pH
• 4. Organize a list of liquids in terms of acid or base strength in relative order with supporting evidence.
• 5. Write the water equilibrium expression. Describe how the water equilibrium varies with pH.

Is this solution acidic or basic?

[H3O + ] = 3.21 x 10 – 4

[OH–] = 3.12 x10 –11

[H3O + ] is greater than [OH–], the solution is acidic.

Acid-Base Definitions

Arrhenius Theory

Bronsted-Lowry Theory

Bronsted-Lowry Acid

A substance that centonate a proton (H+ ions) to some other substance

Bronsted-Lowry Acid

Bronsted-Lowry Base

A substance that can accept a proton (H+ ions)from some other substance

A proton Acceptor

Bronsted-Lowry acid and Bronsted-Lowry Base production must occur simultaneously. You cannot have one without the other.

Acids and Acidic Solutions

Acids

Strength

proticity

Acidity of solutions

acidic solution

(H3o+) > (OH-)

pH < 7.0

neutral solution

(H3O+) = (OH-)

pH = 7.0

Basic solution

(H3o+) > (OH-)

pH > 7.0

[H3o+] [oH-] = 1.0×10-14