A 0.110 M solution of a weak acid (HA) has a pH of 3.26. You may want to reference (Pages 700 - 709) Section 16.7 while completing this problem. Part A Calculate the acid ionization constant (Ka) for the acid. Express your answer using two significant figures.

Ka = 2.76 * 10^-6

Explanation:

The question asks to calculate the acid ionization constant value(Ka) for an acid, given the molarity and pH of the acid in question.

Firstly, we write the dissociation equation for the acid: This can be written as follows:

HA --->H+ + A-

From here, we can write the equilibrium expression of this dissociation equation as:

Ka = [H+][A-]/[HA]

Hence, to calculate the value of Ka, we need to know the value of the concentrations.

We start with the hydrogen ion

Mathematically, pH = -log[H+]

3.26 = -log[H+]

[H+] = -Antilog(3.26)

[H+] = 5.494 * 10^-4 M

We can use the initial, change and equilibrium table I.e ICE to calculate the equilibrium concentration as follows:

HA. [H+]. [A-]

I. 0.11. 0. 0.

C. -x. +x. +x

E. 0.11-0.0005494 0.0005494. 0.0005494

Since 0 + x = 0.0005494, x = 0.0005494

We now calculate the value for the Ka according to the equation below:

Ka = [H+][A-]/[HA]

Ka = (0.0005494)^2/(0.11-0.0005494)

Ka = 0.00000030184/0.1094506

Ka = 2.76 * 10^-6

i'm assuming that it's c.) red bank..