Acid and base behavior can be observed in solvents other than water. One commonly used solvent is dimethyl sulfoxide (DMSO), which can be treated as a monoprotic acid 'HSol.' Just as water can behave either as an acid or a base, so HSol can behave either as a Brønsted–Lowry acid or base. (b) The weak acid HCN has an acid dissociation constant Ka = 1.3 * 10-13 in the solvent HSol. If 0.010 mol of NaCN is dissolved in 1.00 L of HSol, what is the equilibrium concentration of H2Sol + ?

Video transcript

Hi everyone for this problem. It reads consider a hypothetical liquid that undergoes auto ionization similar to water. The auto ionization constant for this hypothetical liquid is one times 10 to the negative 20 suppose you have a 200.51 molar solution of potassium fluoride in this hypothetical liquid. If the dissociation constant of hydrogen fluoride in the hypothetical liquid solvent is 2.3 times 10 to the negative eight, calculate the equilibrium concentration of the following. Okay, so we want to calculate the equilibrium concentration okay, of this. And so let's start off by writing out what is going to be the auto ionization of this hypothetical liquid and auto ionization of this hypothetical liquid is going to be the following. Okay, so essentially we want to calculate the equilibrium concentration of this right here. Okay, so for the hydrogen fluoride, this fluoride and ion is the weak base and we can calculate the KB. Okay, so the KB is going to equal One times 10 to the negative 20, divided by 2.3 times 10 to the -8. These values were given in the problem, we were told that the auto ionization constant for the hypothetical liquid is one times 10 to the negative 20. And that the dissociation constant is 2.3 times 10 to the negativity. So when we divide that out, what we're going to get is 4. 478 Times 10 to the -13. Okay, so now we know what the KB is the auto dissociation constant for the base. So now what we can do is we can write our reaction and create an ice table so that we can continue to solve this problem. So what our reaction is going to be is going to be our weak base which is our and I on here, it's going to react with the hypothetical liquid to produce hydrogen fluoride plus X minus. Okay, so let's go ahead and create an ice table with this. We'll ride out ice on the side. And remember I stands for initial concentration and for our ice table we don't include liquids. Okay, so we're going to just mark that out. We don't include solids or liquids in the ice table or in our equilibrium constant expression. So our Ira represents initial concentration and we're told in the problem that our initial concentration is 0.0 molar solution. Okay, so we put that here 0.51 and then we have zero product. So that means our reaction is moving in the forward direction. And so what that means is our change in concentration for the reactant is going to decrease and our change in concentration for the products is going to increase. So we'll have our changes minus X here and then we have plus X and plus X. We only have one mole of everything. So we only have one X. For each. So let's combine both rose. So 0.51 minus X. This is X. And this is X. So now let's write our KV expression our dissociation. This expression for the base. So it's going to equal the concentration of products over the concentration of reactant. Okay. And so what that means is our product, our concentration of product over concentration of reactant. And we know what the KB value is because we solved it up here. So let's go ahead and plug that in. So what we have is KB which is 4.3478 times 10 to the negative 13 is going to equal. Now let's plug in the equilibrium row of our ice table into these values right here. Okay so we get is equal to X times X over 0.51 minus X. So now let's take a look at the denominator here at this minus X. We can check to see if this X is negligible. The way we're going to do that is we're going to take the initial concentration which is 0.51 and divided by the KB value which is 4.3478 times 10 to the negative 13. If this value is greater than 500 then we can Omit that X. And hear that value is greater than 500. So we're going to admit this X. Because it's negligible. But if that value ended up being less than 500, we would not be able to omit it and we would have to leave it. So let's just simplify this. 4.3478 times 10 to the negative 13 is equal to X squared over 0.51. Okay so now we can go ahead and solve for X. And when we solve for X, what we get is X. Is equal to 1. times 10 to the negative seven is equal to and based off our ice table. This is equal to the concentration of hydro ni um ions. This is what we need. No, excuse me. It's equal to the concentration yes of the hydro knee um ion but also are other products are other an ion. So this is going to be X minus. Okay and so I put hydro ni um ion in. Okay, Excuse me. So yes this X is equal to our concentration of X minus. So now we can solve for our concentration of X. Two X. Plus because in the table it is so this concentration wrote it incorrectly here. Excuse my error. So this is H. Two X. Plus. So this concentration is going to equal one times 10 to the negative 20 over 1.489, one times 10 to the negative seven. Okay and what we get is 6.7 times 10 to the negative 14. So our final answer here is going to be 6. Times 10 to the -14 as our final answer. So let's go ahead and highlight that, and that is it for this problem. I hope this was helpful.