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Question about acid and base
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1)What is the pH of a solution 0.10M in ammonia and 0.151M in ammonium chloride?(Kb for ammonia 1.80x10^-5)2)What volume(in mL) of 0.33M HCl solution must be added to 222mL of 0.27M ammonia solution to give a buffer solution with pH8.5?(Kb for ammonia 1.80x10^-5)3)What is the pH of a solution 0.167M in... 顯示更多 1)What is the pH of a solution 0.10M in ammonia and 0.151M in ammonium chloride?(Kb for ammonia 1.80x10^-5) 2)What volume(in mL) of 0.33M HCl solution must be added to 222mL of 0.27M ammonia solution to give a buffer solution with pH8.5?(Kb for ammonia 1.80x10^-5) 3)What is the pH of a solution 0.167M in acetic acid and 0.189M in soldium acetate? (Ka for acetic acid 1.75X10^-5)
最佳解答:
1) What is the pH of a solution 0.10 M in ammonia and 0.151 M in ammonium chloride? (Kb for ammonia 1.80x10^-5) pOH = pKb - log([NH3]/[NH4^+]) pOH = -log(1.8 x 10^-5) - log(0.1/0.151) = 4.92 pH = 14 - 4.92 = 9.08 2) What volume(in mL) of 0.33M HCl solution must be added to 222mL of 0.27M ammonia solution to give a buffer solution with pH8.5?(Kb for ammonia 1.80x10^-5) Let V mL HCl is added. H^+(aq) + NH3(aq) ≒ NH4^+(aq) Origin no. of moles of NH3 = 0.27 x (222/1000) = 0.05994 mol No. of moles of HCl added = 0.33 x (V/1000) = 0.00033V mol At equilibrium, [NH3]/[NH4+] = (0.05994 - 0.00033V)/0.00033V pOH = pKb - log([NH3/NH4^+]) 14 - pH = -log(Kb) - log([NH3/NH4^+]) 14 - 8.5 = -log(1.8 x 10^-5) - log[(0.05994 - 0.00033V)/0.00033V] log[(0.05994 - 0.00033V)/0.00033V] = -0.755 (0.05994 - 0.00033V)/0.00033V = 10^(-0.755) = 0.176 0.05994 - 0.00033V = 0.000058 0.000388V = 0.05994 V = 154.5 mL 3) What is the pH of a solution 0.167M in acetic acid and 0.189M in soldium acetate? (Ka for acetic acid 1.75X10^-5) pH = pKa - log([CH3COOH]/[CH3COO^-] pH = -log(1.75 x 10^-5) - log(0.167/0.189) = 4.81
其他解答:A215E4A2B88AAE64
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