After 0.050 mole of hydrofluoric acid is dissolved in 200 mL of aqueous solution, it is found that 7.90% of the acid is dissociated at equilibrium. Calculate the equilibrium concentrations of H3O , undissociated hydrofluoric acid (HF), and conjugate weak base of hydrofluoric acid.

[HF] = 0.23M

0.02M = [F⁻]

[H₃O⁺] = 0.02M

Explanation:

The equilibrium of HF with water is:

HF + H₂O ⇄ H₃O⁺ + F⁻

Where F⁻ is the conjugate base of the HF

The initial concentration of HF is:

[HF] = 0.0500mol / 0.200L = 0.25M

Now, in equilibrium, if the 7.90% is dissociated, the undissociated HF is the 100 - 7.90 = 92.1%

0.25M*92.1% = 0.23M = [HF]

The conjugate base will be 0.25M - 0.23M = 0.02M = [F⁻]

And as the H₃O⁺ ion comes from the same equilibrium of F⁻ and is produced in the same proportion:

[H₃O⁺] = 0.02M

1.consider the following balanced equation and answer parts a through. show calculations for each part.  

3 ag(s) + 4 hno3(aq) → 3 agno3(aq) + no(g) + 2 h2o(l)  

a. calculate the number of moles of agno3 produced by the reaction of 3 moles of ag with excess hno3. if the number to the left of the compound represents the number of moles of the compound which are present, then 3 moles of ag react with 4 moles hno3 to produce 3 moles agno3 (aq).

so the answer to a would be 3 moles.

b. calculate the number of moles of agno3 produced by the reaction of 108 grams of ag with excess hno3.  

i'm not sure how to plug in the grams of silver into an equation using moles to find how many moles are produced.

c. calculate the number of grams of agno3 produced by the reaction of 108 grams of ag with excess hno3  

d. calculate the number of moles of no produced as a byproduct of the reaction of 108 grams of ag with excess hno3.  

e. calculate the number of grams of no produced as a byproduct starting with 108 grams of ag and excess hno3.  

explanation: