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posted by  dj kev on 12/1/2008 10:51:46 PM  |  status: Live  

pls HELP!!!

Course Textbook Chapter Problem
General Chemistry chemistry edition 5 author Murray and Fay 14 65
Question Details:
Calculate the {\rm pH} of the following solutions.
(part a) Solution prepared by dissolving 0.19 {\rm g} of sodium oxide in water to give 127.0 {\rm mL} of solution
(part b) Solution prepared by dissolving 1.26 {\rm g} of pure nitric acid in water to give 0.525 {\rm L} of solution.
(part c)
Solution prepared by diluting 39.0 {\rm mL} of 6.6×10−2 {\rm M} {\rm Ba(OH)_2} to a volume of 326.5 {\rm mL}.
(part D)
Solution prepared by mixing equal volumes of 0.21 {\rm M} {\rm HCl} and 0.42 {\rm M} {\rm HNO_3}. (Assume that volumes are additive.)

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posted by Robert Boyle on 12/2/2008 12:38:16 AM  |  status: Live
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dj kev's comment:
"incomplete answer missing partsC&D"
Response Details:
   
                        
                                    Na2O + H2O → 2 NaOH
         Mass of sodium oxide = 0.19 g
         Volume = 127 mL
      Molar mass of sodium oxide = 61.97 g/mol
 
         Moles of sodium oxide = Mass / molar mass
                                             = 0.19 / 61.97
                                             = 0.003 moles
From the equation, 1 mole of sodium oxide gives 2 moles of NaOH.
So moles of NaOH = 2 x 0.003 = 0.006 moles
 
since NaOH dissociates to give 1 OH- ion,
moles of OH-  =  0.006 moles
 
[OH-] =  Moles / volume
           =  0.006 / 0.127
           = 0.047 M
pOH = -log [ OH-]
         = - log [0.047]
         = 1.32
pH = 14 - pOH
      = 14 - 1.32
      = 12.68
 
 
Mass of nitric acid = 1.26 g
Volume = 0.525 L
Molar mass of nitric acid = 63.01 g/mol
 
Molarity =  Mass / ( molar mass x volume)
              =  1.26 / (  63.01 x 0.525 )
              = 0.038 M
Since HNO3  <------>  H+    +   NO3-
  ∴ [H+] =  0.038 M
pH = -log [0.038]
      = 1.42
 
 

 
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posted by dj kev on 12/2/2008 12:47:02 AM  |  status: Live
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Response Details:
thanks however i got no respose to part C and D? will rerate
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posted by dj kev on 12/2/2008 1:28:19 AM  |  status: Live
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Response Details:
(part c)
Solution prepared by diluting 39.0 {\rm mL} of 6.6×10−2 {\rm M} {\rm Ba(OH)_2} to a volume of 326.5 {\rm mL}.
(part D)
Solution prepared by mixing equal volumes of 0.21 {\rm M} {\rm HCl} and 0.42 {\rm M} {\rm HNO_3}. (Assume that volumes are additive.)
(Cramster SME)
posted by Robert Boyle on 12/2/2008 1:49:11 AM  |  status: Live
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dj kev's comment:
"thanks"
Response Details:
(c)
    When diluted, Final volume, V1 = 326.5 mL
                           Molarity , M1 = ?
  Initial volume, V2 = 39.0 mL
   Initial molarity, M2 = 6.6 x 10-2 M
M1 V1  = M2 V2
M1 =  M2 V2 / V1
      = 39 x 6.6 x 10-2 / 326.5
      = 7.88 x 10-3 M
  Barium hydroxide ionizes to give two hydroxide ions
         Ba(OH)2  <-------->   Ba2+  +  2 OH- 
Hence [OH-] = 2 x 7.88 x 10-3
                      = 0.01576 M
pOH = - log [OH-]
         = - log [0.01576]
         = 1.8
pH = 14 - pOH
      = 14 - 1.8
       = 12.2
 
 
(D)
 
equal volumes of 0.21 {\rm M} {\rm HCl} and 0.42 {\rm M} {\rm HNO_3}
Assume that 1 liter of each solution is taken.
          When the volume doubles the concentration becomes halved.
But the volume is doubled for both the solutions.
 
Hence the H+ ion concentrations are  0.21 from HCl and 0.42 M from HNO3.
Total H+ ion concentration = 0.21 + 0.42
                                          = 0.63
 
pH = -log [H+]
      = -log [ 0.63]
      = 0.2006
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