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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 10โ^5 M, thus [H+] = 1.00 x 10โ^5 M ๏ง pH = โlog(1.00 x 10โ^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 โbโ on the periodic table. ๏ท Calculating the pH of strong bases: o Since strong bases 100% ionize, the [OHโ] will be determined by the [strong base] and stoichiometry o Example 1) 1.00 x 10โ^4 M NaOH ๏ง [NaOH] 0 = 1.00 x 10โ^4 M, thus [OHโ] = 1.00 x 10โ^4 M ๏ง pOH = โlog(1.00 x 10โ^4 ) = 4. ๏ง pH = 14 โ 4.000 = 10. o Example 2) 1.00 x 10โ^4 M Ca(OH) 2 ๏ง [Ca(OH) 2 ] 0 = 1.00 x 10โ^4 M, thus [OHโ] = 2(1.00 x 10โ^4 ) = 2.00 x 10โ^4 M ๏ง pOH = โlog(2.00 x 10โ^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 Aโ) ๏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, OHโ^ will always be a product o Use an ICE chart and Kb to determine equilibrium [OHโ] o Use the equilibrium [OHโ] 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 = [๐ป
+] [๐๐๐๐ก๐๐๐ ๐ค๐๐๐ ๐๐๐๐] x 100
o Weak base percent dissociation = [๐๐ป
โ] [๐๐๐๐ก๐๐๐ ๐ค๐๐๐ ๐๐๐ ๐] x 100
5% Dissociation Rule
๏ท When the percent dissociation of a weak acid/base is less than 5%, the value of โโ 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 โxโ because โxโ is much smaller than the initial concentration due to the low percent dissociation. ๏ท How do you know when to ignore โxโ? 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 โxโ. o On the AP exam, you may ignore โxโ, but make sure to state why you are allowed to ignore โxโ.
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 pHโs 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) ๏ 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 โ^ is neutral because it came from a strong acid
๏ท Strong acid + weak base = ACIDIC salt o Example) HNO 3 (aq) + NH 3 (aq) ๏ NH 4 NO 3 (aq) Strong Weak Acidic Salt ๏ง NO 3 โ^ 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 ๏ 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 โ^ is basic because it came from a weak acid and hydrolyzes water according to the equation: ๏ง C 2 H 3 O 2 โ(aq) + H 2 O(l) โ HC 2 H 3 O 2 (aq) + OHโ(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
- Calculate the pH of a 0.100 M aqueous solution of hypochlorous acid, HOCl. (Ka of HOCl = 3.5 x 10-8).
- 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).
- Calculate the percent dissociation of 0.100 M acetic acid, HC 2 H 3 O 2 (Ka = 1.8 x 10-5).
- Calculate the percent dissociation of 3.1 x 10โ^4 M carbonic acid, H 2 CO 3 (Ka = 4.4 x 10โ^7 ).
- If a 0.375M weak acid (HA) dissociates 2.5% when put in water, what is the acidโs Ka value?
- 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.
