A 0.148 M solution of a monoprotic acid has a percent ionization ...

Question

A 0.148 M solution of a monoprotic acid has a percent ionization 
of 1.55%. Determine the acid ionization constant (Ka) for the 
acid.

Accepted Answer

Hey everyone we're asked what is the acid ionization constant? For a 0.135 moller mono product acid with a 1.76% ionization first. Let's go ahead and use our percent ionization in order to determine our concentration of hydrogen ions. We know that our percent ionization is equivalent to the concentration of hydrogen ions divided by the concentration of our acid multiplied by 100%. Since we are in percentages Plugging in our values we get 1.76% equals the concentration of hydrogen ions divided by 0.135 moller Multiplied by 100%,, solving for our concentration of hydro ni um ions. We end up with 1.76 divided by Multiplied by 0.135. This will get us to a concentration Of 2.376 times 10 to the negative 3rd molar. And when we write out our reaction, we know that we have our acid plus our water and this forms our conjugate base of our acid plus our hydro ni um ions. And as we can see right here our concentration of hydro ni um ions is going to be equivalent to the concentration of our conjugate base. So plugging in our values and solving for our asset ionization constant, we know that it's going to be our products over our reactant. So it would be our concentration of hydro knee um ions times the concentration of our conjugate base divided by the concentration of our acid. And we disregard our water since it is a liquid plugging in our values for RK. A, we get 2.376 times 10 to the negative third. And we're going to square this And we're going to divide it by 0.135 molar. This will get us to a k. A. Of 4.26 times 10 to the negative five, which is going to be our final answer. So I hope this made sense and let us know if you have any questions.

A 0.148 M solution of a monoprotic acid has a percent ionization of 1.55%. Determine the acid ionization constant (Ka) for the acid.

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