What is the pH of an aqueous solution made by combining 33.97 mL of a 0.3687 M a

Question

What is the pH of an aqueous solution made by combining 33.97 mL of a 0.3687 M ammonium chloride with 42.50 mL of a 0.3589 M solution of ammonia? 82 Marks: 1 What is the pH of an aqueous solution made by combining 43.71 mL of a 0.4187 M sodium acetate with 42.22 mL of a 0.3422 M solution of acetic acid to which 4.849 mL of a 0.0684 M solution of NaOH was added? 83 Marks: 1 84 What is the pH of an aqueous solution made by combining 43.17 mL of a 0.4260 M sodium acetate with 36.00 mL of a 0.3565 M solution of acetic acid? Marks: 1 Answer 85 What is the pH of an aqueous solution made by combining 14.07 mL of a 0.1959 M hydrochloric acid with 36.07 mL of a 0.3924 M solution of ammonia? Marks: 1 Answer 86 Marks: 1 In a titration of 42.91 mL of 0.3566 M ammonia with 0.4084 M aqueous nitric acid, what is the pH of the solution when 5.00 mL of the acid have been added? 87 Marks: 1 In a titration of 35.40 mL of 0.3532 M ammonia with 0.3876 M aqueous nitric acid, what is the pH of the solution when 35.40 mL have been added? 10.00 mL of the acid Answer In a titration of 42.15 mL of 0.3520 M ammonia with 0.3520 M aqueous nitric acid, what is the pH of the solution when 42.15 mL of the acid have been Marks: 1 added? Answer

Explanation / Answer

82.
     no of moles of NH4Cl   = molarity * volume in L
                            = 0.3687*0.03397   = 0.0125 moles
    no of moles of NH3      = molarity *volume in L
                            = 0.3589*0.0425 = 0.0153 moles
POH = Pkb + log [NH4Cl]/[NH3]
      = 4.75 + log0.0125/0.0153
       = 4.75 -0.08779   = 4.662
   PH = 14-POH
       = 14-4.662 = 9.34
83.no of moles of CH3COONa = molarity * volume in L
                             = 0.4187*0.04371   = 0.0183 moles
no of moles of CH3COOH     = molarity *volume in L
                             = 0.3422*0.04222   = 0.01445moles
no of moles of NaOH = molarity * volume in L
                      = 0.0684*0.004849 = 0.00033moles
no of moles of CH3COONa after addition of 0.00033 moles of NaOH = 0.0183+0.00033 = 0.01863moles
no of moles of CH3COONa after addition of 0.00033 moles of NaOH = 0.01445-0.00033 = 0.01412 moles
PH = Pka + log[CH3COONa]/[CH3COOH]
     = 4.75 + log0.01863/0.01412
     = 4.75 +0.1203   = 4.87
84.no of moles of CH3COONa = molarity * volume in L
                             = 0.4260*0.04371   = 0.0186 moles
no of moles of CH3COOH     = molarity *volume in L
                             = 0.3565*0.036   = 0.012834moles
PH = Pka + log[CH3COONa]/[CH3COOH]
     = 4.75 + log0.0186/0.012834
     = 4.75 +0.1611 = 4.9111

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