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Date Posted: 7/24/2008 11:39:05 AM  Status: Live
pH need answer fast have a quiz
Course Textbook Chapter Problem
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Question Details:
A student titrates 10.0 mL of 0.25 M HNO3 with 0.10 M KOH. What is the pH of the solution after 9.00 mL of the base have been added?
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Date Posted: 7/24/2008 9:59:32 PM  Status: Live
Asker's Rating: Helpful   
Response:
 The neutralization reaction of the HNO3 ,KOH is
 HNO3 +  KOH ------------> KNO3 + H2O
 The resultiong molarity of the solution when 9 mL of the 0.10M KOH was added to the 10 mL of 0.25M HNO3,Thus
   Solution molarity         =  M1V1 + M2V2  / V1 + V2
                                 M = 0.25M * 10mL  + 0.10M * 9mL / 10mL + 9mL
                                      = 0.1789 M
 Therefore pH of the solution = -log{H+]
                                            = - log(0.1789)
                                            =  -log (17.89 * 10-2)
                                            =   0.7474
 

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Date Posted: 7/24/2008 10:48:31 PM  Status: Live
Asker's Rating: Helpful   
Response:
When an acid and base react the resultant molarity has the formula
     R.M = (V1M1 - V2M2)/(V1 +V2)
     V1 = volume of acid HNO3                  V2 = volume of HNO3
     M1 = molarity of HNO3                        M2 = molarity of HNO3
   R.M = (10 x 0.25 - 9 x 0.1)/19
           = (2.5 - 0.9)/19
          = 0.08421
0.08421M is the resultant molarity of acid
    as it is a monobasic acid
  [H+] = 0.08421
pH = -log[H+] 
      = -log0.08421
     = 1.07463 



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