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Lecture Notes Week 04

# Lecture Notes Week 04 - WEEK 4 POLYPROTIC ACIDS(Sect 17.2...

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1 – WEEK 4 – POLYPROTIC ACIDS (Sect. 17.2) e.g., H 2 CO 3 (aq) + H 2 O(l) H 3 O + (aq) + HCO 3 (aq) HCO 3 (aq) + H 2 O(l) H 3 O + (aq) + CO 3 2– (aq) From tables: K a1 (H 2 CO 3 ) = 4.45 × 10 –7 K a2 (HCO 3 ) = 4.69 × 10 –11 Question : What is the pH of a 0.500 M H 2 CO 3 solution and what is [CO 3 2– ]? Strategy : K a1 >> K a2 , therefore ________________________________________. Determine pH based on _____________________________________: [HCO 3 ] = [H 3 O + ] = = 4.72×10 –4 ; 5% rule: 0.09% (assumption valid) pH = 3.326 Determine [CO 3 2– ] based on ______________________________: HCO 3 (aq) + H 2 O(l) H 3 O + (aq) + CO 3 2– (aq) @ equil: 4.72×10 –4 x x x x = {4.69×10 –11 × 4.72×10 –4 } = [CO 3 2– ] [CO 3 2– ] = 1.49×10 –7 M Weak BASES (Section 17.3) MAIN TYPES: (i) Ammonia & its Organic Derivatives NH 3 (aq) + H 2 O(l) NH 4 + (aq) + OH (aq) e.g., CH 3 NH 2 (aq) + H 2 O(l) CH 3 NH 3 + (aq) + OH (aq)

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2 other organic derivatives of ammonia include: (CH 3 ) 2 NH & (CH 3 ) 3 N pyridine (C 5 H 5 N), aniline (C 6 H 5 NH 2 ) hydroxylamine (HONH 2 ) In general : B(aq) + H 2 O(l) BH + (aq) + OH (aq); K b (B) Note: If N is bonded to three things, this leaves its lone pair of electrons available to act as a base (i.e, accept an H + ion). (ii) Conjugate Bases of Weak Acids e.g., acetate, fluoride, nitrite, etc. i.e., A (aq) + H 2 O(l) HA(aq) + OH (aq); K b (A ) e.g., F (aq) + H 2 O(l) HF(aq) + OH (aq); K b (F ) = 1.5×10 -11 MATHEMATICAL TREATMENT OF WEAK BASES (same path as for Weak Acids) B(aq) + H 2 O(l) BH + (aq) + OH (aq) [ I nitial] C b 0 0 [ C hange] - y + y + y [ E quilib] C b - y y y y y - - C ] B [ ] OH ][ BH [ K B 2 b = = + , where y = [OH ] = [BH + ] As before, we can rearrange this into: ) - (C K y ] [OH OR K C B b b 2 B y y y - = = + = N NH 2
3 EXAMPLE 1 : For dimethylamine, (CH 3 ) 2 NH, the pH of a 2.6 × 10 –2 M solution is 11.56.

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Lecture Notes Week 04 - WEEK 4 POLYPROTIC ACIDS(Sect 17.2...

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