Question

When all concentrations at equilibrium are known a numerical value for the constant can be calculated. Give the value for the Keq given the following concentrations:

[HF] = 0.030 M, [H3O+1] = 0.020 M, [F-1] = 0.020 M

HF + H2O ⇔ F-1 + H3O+1

[F-1][H3O+1] [
][ ]

Keq = ––––––––––– = ––––––––––––––––––––––––––––––––––––––––––––––––– =

[HF] [ ]


a. HF b. H2O C. F-1 d. H3O+1 e.0.020 M f. 0.030 M G. 0.013


Units are not usually used with equilibrium constants. This Keq can also be called a Ka since it is for an acid.

Answer 1

Keq = 0.013

Step-by-step explanation:

Keq is the ratio of the concentrations of the products to the concentration of the reactants.

For any reaction aA+bB->cC+dD (a,b,c,d are coefficients),
`K_(eq)=([C]^c[D]^d)/([A]^a[B]^b)`.

Furthermore, pure solids and liquids are omitted from these Keq expressions; the only things included in Keq expressions are aqueous or gaseous substances.

Since the coefficients for all reactants and products in the given reaction is 1 and water is omitted because it is a pure liquid,
`K_(eq)=([F^-][H_3O^+])/([HF])`.

Plugging in the concentrations gives
`K_(eq)=(0.020*0.020)/(0.030)=(0.020^2)/(0.030)=(0.00040)/(0.030)=0.013`.