Kemi

puffer hjælp

01. januar 2014 af KIKULI (Slettet) - Niveau: A-niveau

Har brug for hjælp med den vedhæftede opgave

Vedhæftet fil: Udklip 2.PNG

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Svar #1
01. januar 2014 af mathon

a)
          pufferligningen giver
                                                pH = pKs + log(nb/ns)
 

                                                nb      
                                               ----- = 10pH-pKs
                                                ns           

.

                                                Vb•cb           
                                               --------- = 10pH-pKs
                                                Vs•cs                 

.

                                                Vb           
                                               ----- = 107,2-7,4 = 0,63096         da   cb = cs = 0,5 M
                                                Vs                   

.

                                                       Vb           
                                               ---------------- = 0,63096       
                                                (0,5 L) - Vb       


                                              Vb = V(Na2HPO4) = 0,1934 L = 193,4 mL

                                              Vs = V(NaH2PO4) = (500 mL) - (193,4 mL) = 306,6 mL
                                          


Svar #2
01. januar 2014 af KIKULI (Slettet)

                                                                           Vb           
    skal det ikke være                                           ----- = 107,4-7,2, da pH er 7,4 og pKs er 7,2?
                                                                            Vs                   


Brugbart svar (0)

Svar #3
01. januar 2014 af mathon

korrektion:

  pufferligningen giver

                                                Vb           
                                               ----- = 107,4-7,2 = 0,63096         da   cb = cs = 0,5 M
                                                Vs                   

.

                                                       Vb           
                                               ---------------- = 1,58489    
                                                (0,5 L) - Vb       


                                              Vb = V(Na2HPO4) = 0,1934 L = 193,4 mL

  pufferligningen giver
                                                pH = pKs + log(nb/ns)
 

                                                nb      
                                               ----- = 10pH-pKs
                                                ns           

.

                                                Vb•cb           
                                               --------- = 10pH-pKs
                                                Vs•cs                 

.

                                                Vb           
                                               ----- = 107,4-7,2         da   cb = cs = 0,5 M
                                                Vs                   

.

                                                       Vb           
                                               ---------------- = 1,58489     
                                                (0,5 L) - Vb       


                                              Vb = V(Na2HPO4) = 0,3066 L = 306,6 mL

                                              Vs = V(NaH2PO4) = (500 mL) - (306,6 mL) = 193,4 mL
                                          

                                         


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Svar #4
01. januar 2014 af mathon

b)
       tilsætning af NaOH

                                         H2PO4- (aq)    +   OH- (aq)   --->   HPO42- (aq)    +   H2O (l)
       nye molariteter            0,5-x                   x                    0,5+x

                                                0,5+x     
                                               --------- = 108,4-7,2       0 < x < 0,5
                                                0,5-x        

                                                x = 0,4406 mol

                                                m(NaOH)tilsat = (0,4406 mol) · (40,00 g/mol) = 17,624 g NaOH

                      
                                           

                         


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Svar #5
01. januar 2014 af mathon

oprydning i korrektionsrod hvor øverste taste-forkerte del ikke blev slettet:
 

  pufferligningen giver
                                                pH = pKs + log(nb/ns)
 

                                                nb      
                                               ----- = 10pH-pKs
                                                ns           

.

                                                Vb•cb           
                                               --------- = 10pH-pKs
                                                Vs•cs                 

.

                                                Vb           
                                               ----- = 107,4-7,2         da   cb = cs = 0,5 M
                                                Vs                   

.

                                                       Vb           
                                               ---------------- = 1,58489     
                                                (0,5 L) - Vb       


                                              Vb = V(Na2HPO4) = 0,3066 L = 306,6 mL

                                              Vs = V(NaH2PO4) = (500 mL) - (306,6 mL) = 193,4 mL
                                          

                                         


Svar #6
02. januar 2014 af KIKULI (Slettet)

Tusind tak for hjælpen 


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