Rumfang - pH regning

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#1

Har kigget på det, men med mine tal virker det fuldstændig ulogisk og giver en masse mærklige tal! Help??

n(syre) = c * V = 0,100 M * 100 mL = 10 mmol

n(base) = c * V = 0,100 M * V

De reagerer 1:1, så:

n(slut) = (0,100 M * V) - 10 mmol

V(tot) = 100 mL + V

c(slut, base) = n(slut) / V(tot)

pH = 14 - pOH = 14 - [-log (nslut / V(tot) ] = 14 + log[((0,100 M * V) - 10 mmol) / (100 mL + V)]

Så løser du for V.

      
               K_s(HF) = 10^-pK_s = 10^-3,14 = 7,2·10^-4

#4

Hvor får du -3,14 fra? Og hvorfor udregner du K_s for HF, når den allerede er opgivet?

HF
              er en middelstærk syre
                                                               pH = 4 = -log((-K_s+√(K_s²+4K_s·c_s))/2)

                                                                        10^-4 = (-K_s+√(K_s²+4K_s·c_s))/2

                                                                        2·10^-4 = -K_s+√(K_s²+4K_s·c_s)

                                                                        2·10^-4 = -10^-3,14+√(10^-6,28 + 4·10^-3,14·c_s)

                    hvoraf 
                                                                        c_s = 1,138·10^-4 M

de 100mL 0,100 M flussyre indeholder 0,01 mol [H₃O⁺]

hvoraf
                                                                                                             _{NaOH                          HF}
                                                    (0,01 mol) - V·(0,100 mol/L) = V·(1,138·10^-4 mol/L)       da HF : NaOH
                                                                                                                                                1   :   1
          som giver                   
                                                                V = 0,09987 L NaOH = 99,87 mL NaOH

     pKs = 3,14

korrektion til #6

HF
              er en middelstærk syre
                                                               pH = 4 = -log((-K_s+√(K_s²+4K_s·c_s))/2)

                                                                        10^-4 = (-K_s+√(K_s²+4K_s·c_s))/2

                                                                        2·10^-4 = -K_s+√(K_s²+4K_s·c_s)

                                                                        2·10^-4 = -10^-3,14+√(10^-6,28 + 4·10^-3,14·c_s)

                    hvoraf 
                                                                        c_s = 1,138·10^-4 M

de 100mL 0,100 M flussyre indeholder 0,01 mol [H₃O⁺]

hvoraf
                                                                                                             _{NaOH                          HF}
                                                    (0,01 mol) - V·(0,100 mol/L) = (V+0,100 L)·(1,138·10^-4 mol/L)      

                                                                                                                                         da HF : NaOH
                                                                                                                                               1   :   1
          som giver                   
                                                                V = 0,987623 L NaOH = 987,623 mL NaOH

Hvor får du -3.14 og -6,28 fra?

2·10-4 = -10^-3,14+√(10^-6,28 + 4·10^-3,14·c_s)

Desuden, når jeg udregner c_s får jeg ikke 1,138*10^-4 M. Hvordan får du det?

Jeg har også prøvet at udregne (0,01 mol) - V·(0,100 mol/L) = (V+0,100 L)·(1,138·10^-4 mol/L), men jeg får heller ikke 0,987623 L.

desværre fejltast
som korrigeres til

                                                                     _{NaOH                          HF}
                               (0,01 mol) - V·(0,100 mol/L) = (V+0,100 L)·(1,138·10^-4 mol/L)    

                                             V = 0,099773 L = 99,773 mL