An acid solution with a concentration of 0.500 M has a pH = 3.21. What is the Ka of the acid? (LO 16.8) (a) 1.2 * 10-5 (b) 1.7 * 10-6 (c) 7.6 * 10-7 (d) 5.4 * 10-3

Video transcript

Hello. In this problem we are told the ph of 8.75 moller acid solution is 4.67. Brass determined the acid dissociation constant for the acid. Let's begin by writing. Then the equation that describes the dissociation of a weak acid. So we have our weak acid. H. A 12 react with water. The acid will donate a proton to water reform. Hydro mines and the conjugate base of our acid. Making use of this equation that's now generate an ice table. So initially we have .75 Mueller acid. We ignore the water because it's a pure liquid and initially we have none of our products present. Our change then is minus X plus X and plus X. We combined initial and change to come up with the equilibrium recall then that we can use the ph to find the concentration of hydrogen mines equilibrium. So our hydrogen mind concentration equal to to the negative ph And this works out to 2.13, 8 times 10 - Mueller. This is equal to X. In the table. Which also tells us in the concentration of our contract base. So writing are acid dissociation constant, expression. We have the concentration of our products over the concentration of our reactant. And this then given ice table is equal to x times x over 0.75. Kind of sex plugging in the value that we found for X. You get 2.138 times 10 -5 squared. However 0.75 - Our value for X. And so our acid dissociation constant Works out to 6.1 Times 10 to the -10. And so this value for the association constant for our weak acid corresponds to answer D. Thanks for watching. Hope this helps.