Pyridinium bromide 1C5H5NHBr2 is a strong electrolyte that dissoc...

Question

Pyridinium bromide 1C5H5NHBr2 is a strong electrolyte that dissociates completely into C5H5NH+ and Br-. An aqueous solution of pyridinium bromide has a pH of 2.95. (c) A solution of pyridinium bromide has a pH of 2.95. What is the concentration of the pyridinium cation at equilibrium, in units of molarity?

Accepted Answer

Hi everyone this problem reads the ph of a solution of nitrous acid is 4.56. What is the concentration of nitrous acid in malaria? T Okay so we want to know the concentration of nitrous acid. So let's go ahead and write out our equilibrium expression. So we have our nitrous acid dissociates into hydro ni um ions plus our nitrate our nitrite ion. Excuse me. Okay so with that with this reaction we can write our K. Expression, our acid dissociation expression which is K. A. Is equal to our products over our reactant. Okay and in this problem we are actually given the value of K. A. So that value is 4.6 times 10 to the -4. And in the problem we are asked to find the concentration of the of the nitrous acid. But first what we can do is because we know the P. H. Before we even start solving for the concentration of nitrous acid. Because we know the P. H. We can first solve for our concentration of hydro ni um ion. Okay and that's because we know the concentration of hydro knee um ion is going to equal 10 to the neck negative P. H. And we were told what the ph is and the problem so we can go ahead and plug that in. So 10 to the negative 4.56. Okay so that means our concentration of hydro ni um ion is going to equal 2.7542 times 10 to the negative five. And our concentration of hydrogen ions is going to equal our concentration of nitrite ion. So this number is the same for both. So now that we know that we can now solve for our concentration of nitrous acid. Okay, so let's go ahead and substitute in what some values for our K. Expression. So we know that our concentration of hydro ni um and nitrite ion are the same. So we're going to have 2. times 10 to the negative five squared. So we're going to have this over our concentration of of nitrous acid and our concentration of nitrous acid is going to be x minus our concentration of hydro ni. Um So X. This is the same thing as X. So it's going to be excuse me. No we're going to do x minus the value we solve for so X minus 2.7542 times 10 to the negative five. So this X. Here represents our concentration of nitrous acid. So this is equal to R. K. A. Which is 4.6 times 10 to the negative four. So when we solve for X, that is going to give us our concentration of nitrous acid and this is our K. A. O. K. R K. Is equal to our products over our reactant and we already know the value of K. A. So now we're solving for X. So let's go ahead and cross multiply. Okay so we're going to get 4. times 10 to the negative four times x minus 2.7542 times 10 to the negative five is going to equal 2.7542 times 10 to the negative five squared. So we just cross multiply to get rid of the fractions. Okay, so now for the left side we're going to foil This 4.6 times 10 to the -4. Okay, so let's go ahead and do that. So we get when we foil it out we get 4.6 times 10 to the negative four. X minus 1. times 10 to the negative eight Is equal to 7.585. times 10 to the negative 10. Okay, so now we're going to add 1. times 10 to the negative eight to both sides. Okay. And when we do that we get 4.6 Times 10 to the -4. X is equal to 1. 696 times 10 to the negative eight. So remember we're solving for X here. So let's go ahead and divide both sides by 4.6 times to the negative four. Okay. And so we get X is equal to 2.9190 times 10 to the negative five. Remember we said that this X based off of our K expression up above is going to equal our concentration for our nitrous acid. Okay, so that means our final answer. Our concentration of nitrous acid and malaria T is 2.9 times 10 to the - molar. Okay, so this is our final answer, and that's it for this problem. I hope this was helpful.

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