KF is a strong electrolyte, and HF is a weak electrolyte. How is ...

Question

KF is a strong electrolyte, and HF is a weak electrolyte. How is the solution of KF different from that of HF?

Accepted Answer

Welcome back everyone. Potassium sulfate K two S 03 is a strong electrolyte since only our potassium Canion and our sulfate anions are present when it is dissolved in water. While sulfurous acid H two S 03 is a weak electrolyte. Since its acre solution consists of mostly water and our sulfurous acid molecules with a few hydro and hydrogen sulfide ions were asked whether this statement is a true statement. So let's consider our first compound mentioned potassium sulfide K two S 03 based on the chemical formula containing metals being our potassium atoms and nonmetals being sulfur and oxygen within our formula. We have an ionic compound. And so because we have an ionic compound, we ultimately have a salt and recall that ionic compounds will based on their solubility, dissolve in water into their ions. In this case, we're going to recall the solubility of sulfates in which we recall that all sulfates are going to be soluble. So all sulfates are soluble. With the exception of lead sulfates, barium sulfates, calcium sulfates and strontium sulfates. In this case, we have a potassium sulfate which will be soluble fully. So we're going to use a regular reaction arrow pointing towards the right where we would dissolve into our potassium K plus ions which are ions. And then we have our second product which will be our sulfate anion S +032 minus note that I'm giving the label A Q which represents that each of our substances on both the reactant and product side are aqueous and soluble in water. Now, because our ionic salts, potassium sulfite fully dissolved into its ions as we indicated by our forward reaction arrow. This does confirm that therefore, our potassium sulfide and that's K two S 03 is a strong electrolyte. And that is because in solution, we have its ions, potassium and sulfide. Within our solution of water, we will no longer have our potassium sulfite molecule. So the first half of our statement definitely checks out as correct. Now, we're going to consider our second compound mentioned which is sulfurous acid H two S 03. Now, recalling upon our memorized list of strong acids and weak or strong acids and strong bases, we're going to recognize that sulfurous acid is not on our list of memorized strong acids and it's certainly not a base. So therefore, sulfurous acid is considered a weak acid. We're going to recall that whenever we have a weak acid or weak base, our substance will be in equilibrium in a hydrolysis reaction where we're going to need to include water as our second reactant. So our weak acid is acre and in this case, and our acid is going to donate a proton. So one of our H plus two water, which will be the proton acceptor. And specifically, we want to show our arrow pointing to our oxygen where the proton donated from our sulfurous acid forms a bond with one of the lone pairs on oxygen. This will result in our first product formed in solution, which will be hydro. And then we will also have as a result of our acid donating a proton. Our conjugate base being our hydrogen sulfide anion H S 03 minus note that our equilibrium arrow is important because it differentiates from our first substance which fully dissolved into its ions. Whereas equilibrium shows that we will have mostly a higher concentration of our sulfurous acid molecules and water molecules versus its dissolved ions hydro and hydrogen sulfide. So ultimately, we can confirm that for our second compound, sulfurous acid, we have molecules plus ions in solution. Whereas in our first solution of potassium sulfide, we had only ions in solution. Now because we have molecules and ions in our solution of sulfurous acid, we would say therefore, sulfurous acid H two S 03 is a weak electrolyte. So our final answer to complete this example is going to be that our given statement is a true statement. So we'll say yes, this is sorry, this says yes, this is a true statement. And this would be our final answer. I hope that this made sense and let us know if you have any questions.

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KF is a strong electrolyte, and HF is a weak electrolyte. How is the solution of KF different from that of HF?

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