1. 14.3 pH Scale 14.4 pH of Strong Acids

2. The pH Scale & pH Values of Some Common Substances
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pH Range
pH = 7; neutral
pH > 7; basic
Higher the pH, more basic.
pH < 7; acidic
Lower the pH, more acidic.
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14.3 – The pH Scale
pH = –log[H+]
pH changes by 1 for every power of 10 change in [H+].
pH is a quantitative description of solution acidity.
pH decreases as [H+] increases.
Significant figures:
The number of decimal places in the log is equal to the number of significant figures in the original number.
Example: If [H+] = M or 1.0×10-2M
pH =
2.00
Two significant figures!
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14.3 – Calculating pH
EXERCISE!
Calculate the pH for each of the following solutions.
1.0 × 10–4 M H+
pH = -log(1.0 × 10–4) = 4.00
Is this answer reasonable?
Yes! Because an acid is present in a low concentration –the pH is a little below 7
b) M H+
pH = -log(0.0400) = 1.398
Yes! Because an acid is present in a higher concentration –the pH is even lower
pH = –log[H+]
a) pH = –log[H+] = –log(1.0 × 10–4 M) = 4.00
b) Kw = [H+][OH–] = 1.00 × 10–14 = [H+](0.040 M) = 2.5 × 10–13 M H+
pH = –log[H+] = –log(2.5 × 10–13 M) = 12.60
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6. 14.3 – Calculating concentration from pH
EXERCISE!
The pH of a solution is What is the [H+] for this solution?
pH = –log[H+], so [H+] = the inverse log of –pH
[H+] = 10–pH
[H+] = 10–5.85 = 1.4 × 10–6 M
Is this answer reasonable?
Yes! The solution is acidic (pH below 7), but not very acidic – it has a low [H+]
[H+] = 10^–5.85 = 1.4 × 10–6 M
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7. 14.3 pH and pOH The Acid-Base calculation tool kit:
Kw = [H+][OH–] = 1×10-14
–log Kw = –log[H+] – log[OH–]
pKw = pH + pOH
14.00 = pH + pOH
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14.3 – Calculating pOH
EXERCISE!
Calculate the pOH for each of the following solutions.
1.0 × 10–4 M H+
pOH = –two significant figures
Is this reasonable?
Yes! Because pH = 4, & pH + pOH = 14
b) M OH–
pOH = –three significant figures
Yes! [OH-] is fairly high = low pOH and high pH = a basic solution is present
pH = –log[H+] and pOH = –log[OH–] and = pH + pOH
a) pH = –log[H+] = –log(1.0 × 10–4 M) = 4.00; So, = pOH; pOH = 10.00
b) pOH = –log[OH–] = –log(0.040 M) = 1.40
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9. 14.3 – Calculating concentration from pOH
EXERCISE!
The pH of a solution is What is the [OH–] for this solution?
[H+] = 10–5.85 = 1.4 × 10–6 M
[H+][OH–] = 1×10-14
(1.4 × 10–6 M)[OH–] = 1×10-14
[OH–] = 1×10-14 ÷ 1.4 × 10–6 M = 7.1 × 10–9 M
Or: 14 – pH = pOH; = pOH = 8.15
[OH-] = 10–8.15 = 7.1 × 10–9 M
[OH–] = 7.1 × 10–9 M
pH = –log[H+] and pOH = –log[OH–] and = pH + pOH
14.00 = pOH; pOH = 8.15
[OH–] = 10^–8.15 = 7.1 × 10–9 M
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10. Thinking About Acid–Base Problems
What are the major species in solution?
What is the dominant reaction that will take place?
Is it an equilibrium reaction or a reaction that will go essentially to completion?
React all major species until you are left with an equilibrium reaction.
Solve for the pH if needed.
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11. YES! Because a strong acid is present in a pretty high concentration!
14.4 – Calculating pH of a Strong Acid
CONCEPT CHECK!
What is the pH of an aqueous solution of 2.0 × 10–3 M HCl. What are the major species in solution? H+, Cl–, H2O HCl  H+ + Cl– Also: H2O  H+ + OH– pH will be affected by [H+] Main source of [H+]? What is the pH? pH = -log(2.0 × 10–3 ) = 2.70
2.0 × 10–3 M
Major Species: H+, Cl-, H2O
pH = –log[H+] = –log(2.0 × 10–3 M) = 2.70
pH = 2.70
YES! Because a strong acid is present in a pretty high concentration!
Reasonable answer?
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12. YES! A tiny bit of a strong acid won’t change the pH!
14.4 – Calculating pH of a Strong Acid
CONCEPT CHECK!
Calculate the pH of a 1.5 × 10–11 M solution of HCl. What are the major species in solution? H+, Cl–, H2O HCl  H+ + Cl– Also: H2O  H+ + OH– Main source of [H+]? What is the pH? pH = -log(1.0 × 10–7 ) = 7.00
1.5 × 10–11 M
1 × 10–7 M
NOT!
pH = 10.82
Major Species: H+, Cl-, H2O
pH = –log[H+] = –log(2.0 × 10–3 M) = 2.70
pH = 2.70
YES! A tiny bit of a strong acid won’t change the pH!
Reasonable answer?
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CONCEPT CHECK!
Calculate the pH of a 1.5 × 10–2 M solution of HNO3.
Major Species: H+, NO3-, H2O
The reaction controlling the pH is: H2O + H2O  H3O+(aq) + OH-(aq)
This is because it is the only equilibrium reaction (NO3- is not a base in water). However, the major source for H+ is from the nitric acid, so the pH = -log(1.5 x 10-2) = 1.82.
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Let’s Think About It…
When HNO3 is added to water, a reaction takes place immediately:
HNO3 + H2O H3O+ + NO3–
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Let’s Think About It…
Why is this reaction not likely?
NO3–(aq) + H2O(l) HNO3(aq) + OH–(aq)
Because the K value for HNO3 is very large and HNO3 virtually completely dissociates in solution.
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Let’s Think About It…
What reaction controls the pH?
H2O(l) + H2O(l) H3O+(aq) + OH–(aq)
In aqueous solutions, this reaction is always taking place.
But is water the major contributor of H+ (H3O+)?
pH = 1.82
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