62-Ka-Kb Problems-Ans

62-Ka-Kb Problems-Ans - ACID BASE GCHEM II LAB CONCEPT...

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ACID – BASE GCHEM II LAB CONCEPT PROBLEMS 1. I.D. acid w/circle - conjugate base w/square I.D base w/square - conjugate acid w/circle a) HF(a) + H 2 O(l) H 3 O +1 (a) + F -1 (a) F -1 (a) + H 3 O +1 (a) H 2 O(l) + HF(a) b) NH 3 (a) + H 2 O(l) NH 4 +1 (a) + OH -1 (a) NH 4 +1 (a) + OH -1 (a) H 2 O(l) + NH 3 (a) c) CHO 2 -1 (a) + H 3 O +1 (a) H 2 O(l) + HCHO 2 (a) H 2 O(l) + HCHO 2 (a) H 3 O + (a) + CHO 2 -1 (a) d) HSO 4 (a) + CO 2 (g) SO 4 -3 (a) + HCO 3 (a) HCO 3 (a) + SO 4 -3 (a) CO 2 (g) + HSO 4 (a) 2. Complete pH and pOH table: pH [H + ] pOH [OH - ] Is solution acid or base ? 1.34 4.57 x 10 -2 12.7 1.20 x 10 -13 acid 11.5 3.45 x10 -12 2.54 2.90 x 10 -3 base 4.46 3.47 x 10 -5 9.54 2.88 x 10 –10 acid 12.8 1.51 x 10 -13 1.18 6.64 x 10 –2 base 3. What is the pH and pOH of a 0.0452 M Mg (OH) 2 assume strong base solution for calculation ? Mg(OH 2 ) (aq) Mg +2 (aq) + 2OH -1 (aq) (.0452) (.0452) (2)(.0452) pOH = – log[OH -1 ] pOH = – log[.0904] = – (– 1.043) pOH = 1.04 14.0 = pOH + pH pH = 14.0 – 1.043 = 12.96 pH = 13.0 4. What is pH of solution made by mixing 35.5 mls of 0.0250M NaOH and 78.3 mls of 0.0150M KOH ? NaOH: (.025) (.0355) = 8.875 x 10 -4 M KOH: (.015) (.0783) = 1.174 x 10 –3 M NaOH(a) Na +1 (a) + OH -1 (a) KOH(a) K +1 (a) + OH -1 (a) 8.875 x 10 -4 (a) 8.875 x 10 -4 (a) 8.875 x 10 -4 (a) 1.174 x 10 -3 1.174 x 10
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