Calculate K_a for the anilinium ion ion given that K_b for aniline (C_6H_5NH_2)
ID: 1062644 • Letter: C
Question
Calculate K_a for the anilinium ion ion given that K_b for aniline (C_6H_5NH_2) is 1.5 times 10^-10 b) Calculate K_b for the cyanide ion given that K_a for HCN(aq) 1.5 times 10^-10 Show how the following ions are hydrolyzed in water and tell if the ion would give an acidic or basic solution: i) CH_3CO_2^- ii) NH_4^+ iii) F^- iv) Fe^3+ v) CH_3NH_3^+ Calculate the pH of a buffer solution prepared by mixing 10.0 mL of 0.25 M ammonia and 15.0 mL of 0.15 M in ammonium chloride. Assume the volumes are additive. K_b for ammonia is 1.8 times 10^-5 Calculate the pH of a buffer solution prepared by mixing 15.0 mL of 0.20 M acetic acid and 15.0 mL of 0.20 M sodium acetate. Assume the volumes arc additive. K_a for acetic acid is 1.8 times 10^-5 Give a recipe to prepare an acetate/acetic acid buffer having a pH of 4.90. K_a for acetic acid is 1.8 times 10^-5 Give a recipe to prepare a ammonia/ammonium chloride buffer having a pH of 8.50. K_b for ammonia is 1.8 times 10^-5 Sketch the following titration curves, clearly mark the end points then tell what information can be obtained from each: a) a strong/strong base b) a weak monoprotic acid/strong base c) a weak diprotic acid/strong base A 25.00 mL of an oxalic acid (H_2C_2O_4) solution required 40.00 mL of a 0.09870 M solution of NaOH for complete neutralization to the phenolphthalein end point. What is the molarity of the acid what was the pH of the oxalic acid solution after the addition of 10.00 mL and 30.00 mL of the base. For H_2C_2O_4, K_a1 = 4.2 times 10^-2 and K_a2 = 6.1 times 10^-5Explanation / Answer
4. a. Ka = Kw/kb = 1*10-14/1.5*10-10
= 6.7*10-5
b. kb = Kw/Ka = 1*10-14/1.5*10-10
= 6.7*10-5
5. Anionic hydrolysis to gives basic solution . cationic hydrolysis to gives acidic solution.
CH3COO- + H2O --------> CH3COOH + OH- basic nature
NH4+ + H2O --------> NH3 + H3O+ acidic nature
F- + H2O ---------> HF + OH- basic nature
Fe+3 + 2H2O -------> Fe(OH)2 + 2H+ acidic nature
Ch3NH3+ + H2O -------> Ch3NH2 + H3O+ acidic nature
6. PKb = -logkb
= -log1.8*10-5
= 4.75
POH = PKb + log[NH4Cl]/[NH3]
= 4.75 + log0.15*15/0.25*10
= 4.75-0.04575 = 4.70425
PH = 14-POH
= 14-4.70425 = 9.2957
7. PH = PKa + log[CH3COONa]/[CH3COOH]
= 4.75 + log0.2*15/0.2*15
= 4.75
Related Questions
drjack9650@gmail.com
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.