What is the percent dissociation of 0.187 M weak acid (HA) with a Ka of 4.3x10-7 at 18.9°C? Enter to 2 decimal places.

Explanation:

Let the reaction equation for the dissociation of weak acid is as follows.

Initial:         0.87            0         0

Change:      -x               +x         +x

Equilibrium:  0.87 - x    +x         +x

Hence, expression for the dissociation constant will be as follows.

Now, putting the given values into the above formula as follows.

               x = 0.000283

Hence, at equilibrium the concentration of hydrogen ions is 0.000283.

or,    

Also,  

                      =

                      = 0.00151

And, the percentage of dissociation is = 0.151%

Thus, we can conclude that percent dissociation of given weak acid is 0.151%.

the graph in the attached figure

step-by-step explanation:

we have

using a graphing tool

the graph in the attached figure

the roots are ( x-intercepts) x=-0.5,x=1,x=3

the y-intercept is the point (0,3)

whatvare the choices

step-by-step explanation: