Write the ka expression for an aqueous solution of hydrofluoric acid: (note that either the numerator or denominator may contain more than one chemical species. enter the complete numerator in the top box and the complete denominator in the bottom box. remember to write the hydronium ion out as , and not as ) ka

Ka = ( [H₃O⁺] . [F⁻] ) / [HF]

Explanation:

HF is a weak acid which in water, keeps this equilibrium

HF (aq)  +  H₂O (l)  ⇄  H₃O⁺ (aq)  +  F⁻ (aq)      Ka

2H₂O (l)  ⇄  H₃O⁺ (l)  +  OH⁻ (aq)   Kw

HF is the weak acid

F⁻ is the conjugate stron base

Let's make the expression for K

K = ( [H₃O⁺] . [F⁻] ) / [HF] . [H₂O]

K . [H₂O] = ( [H₃O⁺] . [F⁻] ) / [HF]

K . [H₂O] = Ka

Ka, the acid dissociation constant, includes Kwater.

i think it is true since you can test the answer and can use observations to complete it.

2b2t

explanation:

2b2t