The pH of a 0.010 M aqueous solution of a weak monoprotic acid, HX, is 4.5. What is the value of the acid ionization constant?

pH = -log(H^+) = 4.5

(H^+) = 3.16E-5

.............HX ==> H^+ + X^-
I..........0.10M...0.....0
C............-x.....x.....x
E..........0.10-x...x.....x

x = 3.16E-5. Substitute into Ka expression and solve for Ka.

To find the value of the acid ionization constant (Ka), we need to use the pH value and the concentration of the acid (HX) in the solution. The acid ionization constant is a measure of how much the acid dissociates in water.

First, we need to calculate the concentration of the hydronium ion (H3O+) in the solution using the equation:

pH = -log[H3O+]

Substituting the given pH value into the equation, we get:

4.5 = -log[H3O+]

To solve for [H3O+], we need to convert the pH value from log scale to linear scale by taking the antilog:

[H3O+] = 10^(-pH)

[H3O+] = 10^(-4.5)

[H3O+] ≈ 3.16 x 10^(-5) M

Since HX is a monoprotic acid, we can assume that the concentration of the acid is equal to the concentration of the hydronium ion:

[HX] ≈ [H3O+] ≈ 3.16 x 10^(-5) M

Now, we can calculate the value of the acid ionization constant (Ka) using the equation:

Ka = ([H3O+][X-])/[HX]

Substituting the values we calculated, we get:

Ka = (3.16 x 10^(-5) M)(3.16 x 10^(-5) M)/(0.010 M)

Ka ≈ 9.98 x 10^(-9)

Therefore, the value of the acid ionization constant for the weak monoprotic acid HX is approximately 9.98 x 10^(-9).