Chem 102 Recitation pH, Weak Acid & Base ---Dr. Q. Wang Mar 26, 2015 [H+], [OH-], and pH At 25 C, for any aqueous solutions, [H3O+] and [H+] are the same thing. [H+] × [OH-] = 1.0 × 10-14 -log[H+] = pH, [H+] = 10-pH -log[OH-] = pOH, [OH-] = 10-pOH pH + pOH =14.00 pH < 7.0, acidic; pH > 7.0, basic. [H+], [OH-], and pH [H+] = 10-pH = 10-3.15 = 7.1×10-4 M [H+] × [OH-] = 1.0 × 10-14 [OH-] = (1.0 × 10-14)/ [H+] = (1.0 × 10-14)/ 7.1×10-4 = 1.4 × 10-11 M pH < 7.0 acidic. [H+], [OH-], and pH [H+] × [OH-] = 1.0 × 10-14 [OH-] = (1.0 × 10-14)/ [H+] = (1.0 × 10-14)/ 3.7×10-9 = 2.7 × 10-6 M pH > 7.0 basic. pH=-log[H+] = -log(3.7×10-9) = 8.43 Salt solution Solutions of many salts are neutral in pH. Both cation and anion can’t form weak base or acid, the salt solution will be neutral in pH. Solutions of many other salts are acidic or basic. If the cation can form weak base, the salt solution will be acidic. If the anion can form weak acid, the salt solution will be basic. Salt solution Question 2 – Determine if each salt will form a solution that is acidic, basic, or pH-neutral. a. Al(NO3)2 b. K2CO3 c. NaF d. CaBr2 a. Cation Al3+ can form a weak base Al(OH)3, anion can’t form a weak acid. Hence, the solution of this salt will be acidic. Salt solution Question 2 – Determine if each salt will form a solution that is acidic, basic, or pH-neutral. a. Al(NO3)2 b. K2CO3 c. NaF d. CaBr2 b. Cation K+ can’t form a weak base, anion CO32- can form a weak acid. Hence, the solution of this salt will be basic. c. Cation Na+ can’t form a weak base, anion F- can form a weak acid. Hence, the solution of this salt will be basic. d. ?????? Conjugate acid/base conjugate acid HB (aq) ⇌ H+ (aq) + base B- (aq) B- (aq) + H2O ⇌ HB (aq) + OH- (aq) Ka of HB Kb of B- acid base conjugate KaKb = Kw = 1.0 X 10-14 or pKa + pKb = pKw = 14.00 Conjugate acid/base HNO2 (aq) H 2O(l ) NO2 (aq) H 3O (aq) NO2 (aq) H 2O(l ) HNO2 (aq) OH (aq) Kw 1.0 1014 11 Kb 2 . 2 10 Ka 4.6 104
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