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 answer is 510

step-by-step explanation:

15 • 13 = 195

15 • 12 = 180

15 • 5 = 75

12 • 5 = 60

195 + 180 = 375

375 + 75 = 450

450 + 60 = 510

a negative exponent is translated as follows:

so, we have