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PERCHLORIC ACID: HClO 4 + H 2 O H 3 O + + ClO 4 - HClO 4 H + + ClO4 - H 3 O + = H +. Strong acids are : HBr HCl HNO 3 H 2 SO 4. NaOH Na + + OH - hydroxide of sodium or sodium base. Strong bases are:

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## Strong acids are : HBr HCl HNO 3 H 2 SO 4

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**PERCHLORIC ACID:**HClO4 + H2O H3O+ + ClO4- HClO4 H + + ClO4- H3O+ = H + Strong acids are : HBr HCl HNO3 H2SO4**NaOH Na+ + OH-**hydroxide of sodium or sodium base Strong bases are: LiOH KOH Ca(OH)2 Sr(OH)2 Ba(OH)2**ACETIC ACID:**CH3COOH + H2O H3O+ + CH3COO- this is a weak acid This chemical reaction is displaced to left**In distilled water there are two iones:**H+ and OH- H+→ from ACID and OH-→ from BASES Don’t forget : Kw = (H+ ) x (OH-) = 10 -14 M (H+ ) = 10 -7 = (OH-) = 10 -7**In H2O : HCl 10-3→ (H+ ) = 10-3Kw = (H+ )**x (OH-) = 10 -14Kw = (10-3) x (10-11) = 10 -14 2) In H2O : NaOH 10-4 M →(OH-) = 10-4 Kw = (H+ ) x (OH-) = 10 -14 Kw = (10-10) x (10-4) = 10 -14**The pH of a solution :**pH = -log10[H+] [H+]= 10-ph pOH = -log10[OH-] pH + pOH = 14**pH of frosty orange juice**pH = -log[H+] = -log[2.9 x 10-4] =3.54**HClO** H+ + ClO- .100 - x x x Ka = __x2__ .100 - x 3.5 x 10-8 = __x2__ 100 Weak Acid Example: Calculate the pH of a 0.100 M solution of HClO The x in the denominator can be dropped because Ka/M is less than 10-3. If Ka/M is greater than 10-3 you have to use the quadratic formula to solve the equation. Therefore: 3.5 x 10-9 = x2 x = 5.9 x 10-5 [H+] = 5.9 x 10-5 M[ClO-] = 5.9 x 10-5 M[HClO] = .100 M - 5.9 x 10-5 ~= .100 M pH = -log(5.9 x 10-5) = 4.2**C6H5NH2** C6H5NH3+ + OH- 20 - x x x Kb = __x2__ 20 - x Kb = __x2__ 20 Weak Base Example: What is the concentration of OH- of a .20 molar solution of aniline? The x in the denominator can be dropped because Kb/M is less than 10-3. If Kb/M is greater than 10-3 you have to use the quadratic formula to solve the equation. Therefore: x2 = 8.4 x 10-11 x = 9.2 x 10-6 [OH-] = 9.2 x 10-6

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