A \(0.040 M\) solution of a monoprotic acid is 14 percent ionized. Calculate the ionization constant of the acid.

Short Answer

Expert verified
The ionization constant of the acid is \(Ka = 0.000814\).

Step by step solution

01

Converting Percentage Ionization into Concentration

Firstly, calculate the concentration of the ionized acid. Given that the 0.040 M solution of acid is 14 percent ionized, the concentration of ionized acid ([H+] = [A-]) would be 0.14 * 0.040 M = 0.0056 M.
02

Substitute Values into the Ka Formula

Secondly, substitute the values into the equilibrium constant expression for ionization of a monoprotic acid which is \(Ka = [H+][A-]/[HA]\). Given that [H+] = [A-] and the remaining concentration of HA is ([HA] initial - [H+]), the Ka can be calculated as follows: \(Ka = (0.0056)^2 / (0.040 - 0.0056)\).
03

Calculate the Ionization Constant

Finally, carrying out the computation from Step 2 yields Ka. This gives \(Ka = 0.000814\).

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