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posted by  ^___^ on 11/30/2008 12:22:01 AM  |  status: Live  

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(a) How would you prepare 1.00 L of 0.50 M HNO3 (aq) from "concentrated" (16 M) HNO3 (aq)?

(b) How many milliliters of 0.20 M NaOH (aq) could be neutralized by 100. mL of the diluted solution?

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posted by Werner on 12/1/2008 5:26:25 AM  |  status: Live
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Response Details:
  a) Given data
   M1 = 16 M
   V1 = ?
   M2 = 0.5 M
    V2 = 1.0 L
  For dilution M1V1 = M2V2
           ∴  The volume of  "concentrated" (16 M) HNO3 (aq) required to prepare 0.5 M 1.0 L solution
                                   V1 = M2V2 / M1
                                         = (0.5M)(1.0L) / (16 M)
                                         = 0.03125 L
                                         = 31.25 mL
  Take the 31.25 mL of concentrated acid and dilute it to 1000 mL then you will get 0.5 M solution.
b)  HNO3 (aq)  + NaOH (aq)  --------------> NaNO3(aq)  + H2O(l)
              M1V1/n1 = M2V2/n2
     Here n1 = n2 = 1
        M1V1 = M2V2
               V2 = M1V1/M2
                     = (100 mL)(0.5M)/ (0.2 M)
                     = 250 mL
  The volume of NaOH required to neutralized by 100. mL of the diluted solution = 250 mL
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