Enough of a monoprotic acid is dissolved in water to produce a 0.0169 M solution. The pH of the resulting solution is 2.68. Calculate the Ka for the acid.

Convert pH 2.68 to (H^+) with

pH = -log(H^+).
Then HA --> H^+ + A^-
Ka = (H^+)(A^-)/(HA)
Substitute for (H^+) and (A^-). For (HA) substitute 0.0169-(H^+)

To calculate the Ka (acid dissociation constant) for the acid, we first need to determine the concentration of H+ ions in the solution.

The pH of a solution is defined as the negative logarithm (base 10) of the concentration of H+ ions in moles per liter (M). Mathematically, pH = -log[H+].

Given pH = 2.68, we can calculate the concentration of H+ ions using the equation 10^(-pH):

[H+] = 10^(-pH) = 10^(-2.68)

Now, let's calculate the concentration of H+ ions:

[H+] = 10^(-2.68) = 2.27 x 10^(-3) M

Since this is a monoprotic acid, it dissociates according to the following equation:

HA ⇌ H+ + A-

The initial concentration of HA is equal to the concentration of acid in the solution. In this case, the concentration of acid is given as 0.0169 M. Therefore, the initial concentration of HA is 0.0169 M.

Since the acid dissociates completely, the concentration of H+ ions produced will be equal to the initial concentration of HA. Thus, [H+] = 0.0169 M.

Using this information, we can now calculate the Ka value.

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

Since the acid is monoprotic, the concentration of A- ions is also equal to the concentration of H+ ions, which is 0.0169 M.

Substituting the values into the Ka expression, we get:

Ka = (0.0169)(0.0169)/(0.0169) = 0.0169

Therefore, the Ka value for the monoprotic acid is 0.0169.