# chemistry

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If it requires 54.7mL of a 0.0765 M HCl solution to complete neutralize 25.0 mL of a KOH solution, what is the molarity of the KOH solution.?

• chemistry - ,

M x V = M x V (for 1:1 solutions only and this is a 1:1 solution.

There is another way that works for everything.
Write the equation.
HCl + NaOH ==> NaCl + HOH

mols HCl = M x L = ??
Convert mols HCl to mols NaOH using the coefficients in the balanced equation. Since this is a 1:1 equation (1 mol HCl to 1 mol NaOH), the number of mols NaOH is the same as number of mols HCl.
Then mols NaOH = M x L. YOu know mols, you know L, calculate M.

• chemistry - ,

this is all so confusing

• chemistry - ,

OK. Here is what you do.
Chemistry works by mols. Mols of X react with mols Y to produce mols Z + mols w.

The definitioin of molarity is # mols/Liter of solution. So we go with that definition, solving for
# mols and that give #mols = M x L.
So how many mols HCl to we have?
We have 54.7 mL of 0.0765 M so that is 0.0547 L of 0.0765 M HCl.
Then mols HCl = M x L = 0.0765 x 0.0547 = 0.00418 mols HCl
mols HCl = mols KOH (from the equation I wrote in my first response although I wrote NaOH instead of KOH; KOH is stated in the problem).
mols NaOH = M x L
0.00418 mols NaOH = M KOH x Liters NaOH
0.00418 = M x 0.0250 L (that's 25.0 mL)
Solve for M = 0.00418/0.0250 = M = 0.167 M.
Its a matter of not being intimidated with three or four numbers and an equation.

• chemistry - ,

SEE NOW THAT MAKES MORE SENSE..IS THERE ANY WAY YOU CAN DO THAT WITH THE REST OF MY QUESTIONS. IT GIVES A PERFECT DETAIL FOR ME.

• chemistry - ,

That makes a lot of sense to you because I worked the problem for you. And it doesn't bother me at all to work one problem for you to get the hang of it. But I want you to know how to do the problem yourself. I won't be there to do the work for you when you take a test. Every problem you have posted this evening has been a two or three step problem with numbers that are substituted into one or more equations. The ONLY thing you must do is to substitute the numbers from the problem into the equation and solve the equation. And I've given directions for going from the beginning to the end in steps. To answer your question, no. I will help you do your homework but I won't do it for you.