Let’s assume you have 250 ml of a 1.3M KOH solution and you want to neutralize it with the acid HBr. It takes 350 ml of the HBr solution to do so. What must have been the molarity of the HBr solution? HBr(aq) + KOH(aq) => KBr(aq) + H2O(l)

The stoichiometry is already correct and does not need to be balanced. I'm having a lot of trouble with this because my daughter was sick and I missed this day of class.

It's really quite simple but I would like to correct a minor point on the equation. (I'm picky and that's why I bring it up.). Your statement, "The stoichiometry is already correct and does not need to be balanced." is not quite correct. That statement implies that the equation is not balanced. It would be more correct to say the equation is balanced and the stoichiometry is 1:1.

At any rate, all of these are three step problems.
Step 1. mols of what you have = M x L = mols KOH = 1.3 x 0.250 = ?
Step 2. Then (in this case) mols KOH = mols HBr. In cases where the stoichiometry is not 1:1 it is changed in this step.
Step 3. M = mols/L. You know mols of the HBr, you know L HBr, calculate M HBar.

Note here that you are not "memorizing" or "using" different equations in steps 1 and 3 but just 1. The definition of M = mols/L solution. In step 1 you've rearranged that to mols = M x L and in step 3 you've kept it as is. When I was a student I memorized literally hundreds of formulas. For example, I memorized not 1 but 6 equations for the (P1V1/T1) = (P2V2/T2). Somehow I never caught on that one equation would do it and all I needed to do was to get one under my belt and let algebra take care of the rest. Stupid I guess. At least it certainly was not very bright. :-).

1.3x.250 = .325 Mols KOH
.325 Mols KOH = .325 Mols HBr
Then I'm taking 350ml (.350) and multiplying it by .325 Mols HBr and I get .11375... That doesn't seem quite right.
I really do appreciate your help. I never took chemistry in high school. I joined the military right after high school and then got married and had 2 children. I am finally getting my college education done now and I'm so far behind the learning curb its crazy!

Hi Chris, I believe I am in your Intro Chem class at Alfred State. To do this problem, all you have to do is substitute the equation C(s) V(s)= C(d) V(d)... C stands for the moles (concentration) and the V stands for the mL (volume) if you plug the numbers into the equation---->
(1.3M KOH)(250mL) = (x) (250mL HBr)
which turns into (325)= (350x)
then to get x by itself, you divide 325 by 350 to get an answer of .9M

Your answer is quite correct but let me point out that your solution works ONLY when the reactants are 1:1; if they are not 1:1 you must use the longer process I used. I don't like to use the short answer(as you did) because students then get the idea that it is a general formula and it isn't.

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