If the percentage ionization of an 0.10 m acid is 3.0%, what is its acid dissociation constant? 0.003 9 x 10-5 .0009 .009

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Solution:- percent ionization =

where, is the equilibrium concentration of product ion and c is the initial concentration of the acid.

Let's plug in the values in the formula:

In general, the acid is represented as HA, the ice table is shown as:

I            0.10                                             0             0

C           -X                                               +X             +X

E         (0.10 - X)                                         X               X

where X is the change in concentration that we already have find out using percentage ionization formula.

Let's plug in the values in it:

0.003 is almost negligible as compared to 0.10, so (0.10 - 0.003) could be taken as 0.10.

Second choice is the right one.

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:

false

because it does not have the potential to be entirely used up