0
$\begingroup$

(a) 0.008 M $\ce{Ba(OH)2}$
(b) 0.010 M $\ce{KI}$

The answer is "a" but I do not understand why. The book says that "a" is a strong electrolyte with a total ion concentration of 0.024. How did they calculate the concentration from the information that they gave? And why is it a strong electrolyte based on that calculation. I thought strong electrolytes were (solutes) ions that dissolved completely in a solvent/solution?

$\endgroup$
2
$\begingroup$

Barium hydroxide is a slightly soluble salt. It dissociates in water and gives 3 ions (one ion of barium and two ions of hydroxide) with the concentration of $0.008 \ce{ M}$ for each one. This makes the total concentration of ions to be $0.024 \ce{ M}$. While the total concentration of $\ce{KI}$ is $0.02 \ce{ M}$. So the solution of barium hydroxide is a stronger electrolyte as it has a higher concentration of dissociated ions.

In your book, they treat barium hydroxide as a strong electrolyte, i.e. it dissolved completely in water and dissociates to its constituent ions. This is of course an approximation. While $\ce{KI}$ is a true strong electrolyte; it dissolved completely in water and dissociates to its constituent ions. So, I agree with you with the definition of strong electrolyte, and I hope I clarify the ambiguity.

$\endgroup$
2
$\begingroup$

You are right in thinking that a strong electrolyte is something that dissociates completely in solution.$\ce{Ba(OH)2}$ and $\ce{KI}$ then, being a strong electrolytes, will effectively dissociate into $\ce{Ba^2+}$+ $\ce{2OH-}$ and $\ce{K+}$+ $\ce{I-}$, respectively.

Going into your example, it's a just a matter of multiplication. We can assume that the volume of the solution is 1L for simplicity. Since you begin with 0.008 M $\ce{Ba(OH)2}$, we can say that you begin with 0.008 mol of $\ce{Ba(OH)2}$. In solution this becomes 0.008 mol of $\ce{Ba+}$ and 2(0.008) mol of $\ce{OH-}$, or 0.024 mol of ions in total. Likewise for KI, you are left with 0.010 mol of $\ce{K+}$ and 0.010 mol of $\ce{I-}$, or 0.020 mol of ions in total.

Converting these values back to molarity (by dividing by 1), we get that the concetration of ions in the $\ce{Ba(OH)2}$ solution is 0.024 M, whereas it is only 0.020 M in the $\ce{KI}$ solution.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.