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O solutie 0,1 M de acid propanoic are pH=3. Calculati valoarea Ka a acestui acid.

Răspuns :

Puia68
c,acid= 10⁻¹mol/l ;  pH=3---.> [H3O⁺]= 10⁻³mol/l

CH3-CH2-COOH + H2O⇄H3O⁺ +CH3-CH2COO ⁻

Ka= [H3O][CH3CH2COO⁻]/ [CH3CH2COOH]
     ='H3O⁺]²/Cacid

Ka=(10⁻³)²/10⁻¹=....................calculeaza !!!

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