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pH of Acids and Bases: Strong, Weak, and their Ionization, Study notes of Stoichiometry

A detailed explanation of the concepts of strong and weak acids and bases, their ionization, and the calculation of their ph values using examples. It covers the ionization equations, the use of ice charts and ka or kb values, and the differences between weak acids and bases. It also includes the concept of percent dissociation and the 5% dissociation rule.

What you will learn

  • How to calculate the pH of weak acids and bases?
  • What is the difference between strong and weak acids and bases?
  • How to calculate the pH of strong acids and bases?

Typology: Study notes

2021/2022

Uploaded on 09/12/2022

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pH of Strong Acids
๏‚ท The six strong acids are: HCl, HBr, HI, HNO3, HClO4, and H2SO4.
๏‚ท Calculating the pH of strong acids:
o Since strong acids 100% ionize, the [H+] = the initial [strong acid]
o Example) 1.00 x 10-5 M HCl
๏‚ง [HCl]0 = 1.00 x 10โˆ’5 M, thus [H+] = 1.00 x 10โˆ’5 M
๏‚ง pH = โˆ’log(1.00 x 10โˆ’5) = 5.000
pH of Strong Bases
๏‚ท The eight strong bases are: LiOH, NaOH, KOH, RbOH, CsOH, Ca(OH)2, Sr(OH)2, and
Ba(OH)2. They are Group 1 and 2 hydroxides that form the โ€œbโ€ on the periodic table.
๏‚ท Calculating the pH of strong bases:
o Since strong bases 100% ionize, the [OHโˆ’] will be determined by the [strong
base] and stoichiometry
o Example 1) 1.00 x 10โˆ’4 M NaOH
๏‚ง [NaOH]0 = 1.00 x 10โˆ’4 M, thus [OHโˆ’] = 1.00 x 10โˆ’4 M
๏‚ง pOH = โˆ’log(1.00 x 10โˆ’4) = 4.000
๏‚ง pH = 14 โ€“ 4.000 = 10.000
o Example 2) 1.00 x 10โˆ’4 M Ca(OH)2
๏‚ง [Ca(OH)2]0 = 1.00 x 10โˆ’4 M, thus [OHโˆ’] = 2(1.00 x 10โˆ’4) = 2.00 x 10โˆ’4 M
๏‚ง pOH = โˆ’log(2.00 x 10โˆ’4) = 3.699
๏‚ง pH = 14 โ€“ 3.699 = 10.301
pH of Weak Acids
๏‚ท Since weak acids DO NOT completely ionize, the [H+] DOES NOT EQUAL the initial [weak
acid]
๏‚ท In fact, [H+] < [weak acid]0
๏‚ท An equilibrium will be established between the weak acid molecules (i.e. unionized
molecules, HA) and its ions (H+ and Aโˆ’)
๏‚ท The equilibrium [H+] is used to calculate pH
๏‚ท Write the ionization equation of the weak acid in water, H3O+ will always be a product
๏‚ท Use an ICE chart and Ka to determine equilibrium [H3O+]
๏‚ท Example) Calculate the pH of a 2.00 M solution of nitrous acid, HNO2. (Ka of HNO2 = 6.0
x 10-4)
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pH of Strong Acids

๏‚ท The six strong acids are: HCl, HBr, HI, HNO 3 , HClO 4 , and H 2 SO 4. ๏‚ท Calculating the pH of strong acids: o Since strong acids 100% ionize, the [H+] = the initial [strong acid] o Example) 1.00 x 10-5^ M HCl ๏‚ง [HCl] 0 = 1.00 x 10โˆ’^5 M, thus [H+] = 1.00 x 10โˆ’^5 M ๏‚ง pH = โˆ’log(1.00 x 10โˆ’^5 ) = 5.

pH of Strong Bases

๏‚ท The eight strong bases are: LiOH, NaOH, KOH, RbOH, CsOH, Ca(OH) 2 , Sr(OH) 2 , and Ba(OH) 2. They are Group 1 and 2 hydroxides that form the โ€œbโ€ on the periodic table. ๏‚ท Calculating the pH of strong bases: o Since strong bases 100% ionize, the [OHโˆ’] will be determined by the [strong base] and stoichiometry o Example 1) 1.00 x 10โˆ’^4 M NaOH ๏‚ง [NaOH] 0 = 1.00 x 10โˆ’^4 M, thus [OHโˆ’] = 1.00 x 10โˆ’^4 M ๏‚ง pOH = โˆ’log(1.00 x 10โˆ’^4 ) = 4. ๏‚ง pH = 14 โ€“ 4.000 = 10. o Example 2) 1.00 x 10โˆ’^4 M Ca(OH) 2 ๏‚ง [Ca(OH) 2 ] 0 = 1.00 x 10โˆ’^4 M, thus [OHโˆ’] = 2(1.00 x 10โˆ’^4 ) = 2.00 x 10โˆ’^4 M ๏‚ง pOH = โˆ’log(2.00 x 10โˆ’^4 ) = 3. ๏‚ง pH = 14 โ€“ 3.699 = 10.

pH of Weak Acids

๏‚ท Since weak acids DO NOT completely ionize, the [H+] DOES NOT EQUAL the initial [weak acid] ๏‚ท In fact, [H+] < [weak acid] 0 ๏‚ท An equilibrium will be established between the weak acid molecules (i.e. unionized molecules, HA) and its ions (H+^ and Aโˆ’) ๏‚ท The equilibrium [H+] is used to calculate pH ๏‚ท Write the ionization equation of the weak acid in water, H 3 O+^ will always be a product ๏‚ท Use an ICE chart and Ka to determine equilibrium [H 3 O+] ๏‚ท Example) Calculate the pH of a 2.00 M solution of nitrous acid, HNO 2. (Ka of HNO 2 = 6. x 10-4)

pH of Weak Bases

๏‚ท Same procedure as a weak acid with the following differences: o Write the ionization equation of the weak base in water, OHโˆ’^ will always be a product o Use an ICE chart and Kb to determine equilibrium [OHโˆ’] o Use the equilibrium [OHโˆ’] to calculate pOH o Use pOH to calculate pH ๏‚ท Example) Calculate the pH for a 15.0 M solution of NH 3. (Kb of NH 3 = 1.8 x 10-5)

Percent Dissociation

๏‚ท Strong acids and bases have 100% dissociation ๏‚ท Weak acids and bases will have less than 100% dissociation ๏‚ท Calculating percent dissociation: o Weak acid percent dissociation = [๐ป

+] [๐‘–๐‘›๐‘–๐‘ก๐‘–๐‘Ž๐‘™ ๐‘ค๐‘’๐‘Ž๐‘˜ ๐‘Ž๐‘๐‘–๐‘‘] x 100

o Weak base percent dissociation = [๐‘‚๐ป

โˆ’] [๐‘–๐‘›๐‘–๐‘ก๐‘–๐‘Ž๐‘™ ๐‘ค๐‘’๐‘Ž๐‘˜ ๐‘๐‘Ž๐‘ ๐‘’] x 100

5% Dissociation Rule

๏‚ท When the percent dissociation of a weak acid/base is less than 5%, the value of โ€œโˆ’ xโ€ in a weak acid/base ICE chart can be ignored. Thus, the equilibrium concentration of the weak acid/base will be the same as its initial concentration. This will be the case on the AP exam. On the exam, make sure you note that you are ignoring โ€œxโ€ because โ€œxโ€ is much smaller than the initial concentration due to the low percent dissociation. ๏‚ท How do you know when to ignore โ€œxโ€? o Compare the initial weak acid/base concentration to 100 times Ka (or 100 Kb). If the initial concentration is much larger than 100 Ka (or 100 kb), then ignore โ€œxโ€. o On the AP exam, you may ignore โ€œxโ€, but make sure to state why you are allowed to ignore โ€œxโ€.

Acid Base Properties of Salts

Salts are produced when an acid and base react. Salts are not always neutral. Some hydrolyze with water to produce aqueous solutions with pHโ€™s other than 7.00. To determine if a salt is neutral, acidic or basic you must think about from which acid and base the salt formed.

๏‚ท Strong acid + strong base = NEUTRAL salt o Example) HNO 3 (aq) + NaOH(aq) ๏ƒ  H 2 O(l) + NaNO 3 (aq) Strong Strong Water Neutral Salt ๏‚ง Na+^ is neutral because it came from a strong base ๏‚ง NO 3 โˆ’^ is neutral because it came from a strong acid

๏‚ท Strong acid + weak base = ACIDIC salt o Example) HNO 3 (aq) + NH 3 (aq) ๏ƒ  NH 4 NO 3 (aq) Strong Weak Acidic Salt ๏‚ง NO 3 โˆ’^ is neutral because it came from a strong acid ๏‚ง NH 4 +^ is acidic because it came from a weak base and hydrolyzes water according to the equation: ๏‚ง NH 4 +(aq) + H 2 O(l) โ†” H 3 O+(aq) + NH 3 (aq)

๏‚ท Weak acid + strong base = BASIC salt o Example) HC 2 H 3 O 2 (aq) + NaOH ๏ƒ  H 2 O(l) + NaC 2 H 3 O 2 (aq) Weak Strong Basic Salt ๏‚ง Na+^ is neutral because it came from a strong base ๏‚ง C 2 H 3 O 2 โˆ’^ is basic because it came from a weak acid and hydrolyzes water according to the equation: ๏‚ง C 2 H 3 O 2 โˆ’(aq) + H 2 O(l) โ†” HC 2 H 3 O 2 (aq) + OHโˆ’(aq)

๏‚ท Weak acid + weak base = ??? salt o pH depends on the Ka value of the cation and Kb value of the anion o If Ka is larger, the solution is acidic o If Kb is larger, the solution is basic

Practice

  1. Calculate the pH of a 0.100 M aqueous solution of hypochlorous acid, HOCl. (Ka of HOCl = 3.5 x 10-8).
  2. Calculate the pH of a 1.5 M solution of methylamine, CH 3 NH 2 (Kb of CH 3 NH 2 = 4.38 x 10-4).
  3. Calculate the percent dissociation of 0.100 M acetic acid, HC 2 H 3 O 2 (Ka = 1.8 x 10-5).
  4. Calculate the percent dissociation of 3.1 x 10โˆ’^4 M carbonic acid, H 2 CO 3 (Ka = 4.4 x 10โˆ’^7 ).
  5. If a 0.375M weak acid (HA) dissociates 2.5% when put in water, what is the acidโ€™s Ka value?
  6. Lactic acid (HC 3 H 5 O 3 ) is a waste product that accumulates in muscle tissue during exercise, leading to pain and a feeling of fatigue. In a 0.100 M aqueous solution, lactic acid is 3.7% dissociated. Calculate the value of Ka for this acid.