The pKa of chloroacetic acid is 2.9. If an organic chemist has a 0.10 M aqueous solution of chloroacetic acid at pH 2.4, what percentage of molecules/ions will be in the basic form? 68%

The answer is given below.

Explanation:

We will consider the acid as HA and will set up an ICE table with the equilibrium dissociation of α.

AT pH 2.4 the initial H+ concentration will be 3.98^10-3 M

                      HA →              H+         +      A-

Initial concentration:  0.1    →  3.98 ^10-3       +      0

equilibrium concentration:  0.1(1-α) →   3.98 * 10-3 + 0.1α              0.1α

pKa of chloroacetic acid is 2.9

-log(Ka) = 2.9

Ka = 1.26 * 10-3

From the equation, Ka = [H+] * [A-] / [HA]

1.26 * 10-3 = (3.98 * 10-3 + 0.1α )* 0.1α / 0.1(1-α)      

Since α<<1, we assume 1-α = 1

Solving the equation, we have: α = 0.094

Since this is the fraction of acid that has dissociated, we can say that % of base form = 100 * α= 9.4%

sorry idk the answer well to you i have a explaination

explanation: a formula which shows the arrangement of atoms in the molecule of a compound