Please help me w/ chem!!!!?

Question

i don't understand chemistry at all and i have this problem to do for homework. I really want to learn how to do this but i just can't figure it out. i did get part of it but then after that i'm lost. Can anyone help?

Here's the problem:

What volume of each of the following bases will react completely with 25.00 ml of 0.200 M HCl?
a. 0.100 M NaOH
b. 0.0500 M Ba(OH)2

I set up the first part (involving base a) HCl+Na(OH)-----> H(OH) + NaCl

What i did so far is i changed the vol of HCl to .025 L. I also multiplied .025 L HCl by .200 Mol/L = .005 Mol

Did i do it right so far? Please help me for the rest b/c i can't get it. Thank you so much!!

Answer 1

You're on the right track!

Finding the number of moles of HCl you have was definitely the first thing you need to do. Now, you need to figure out how many moles of NaOH and Ba(OH)2 will react with that amount of HCl and then convert back into liters.

So...for part a), 1 mole of HCl will react with 1 mole of NaOH (the way to know this is that you're trying to get equal numbers of H's and OH's that can combine to form HOH = h2o = water). So since you have .005 mol of HCl, you need .005 mol of NaOH. Now you need to find out how many liters of .1 molar solution that is. Set up an equation like this:

.005 mol
____________ = .1 M
x L

and solve to get x = .05 L, or 50 mL.

For part b), you do the same thing except that you only need half as many moles of Ba(OH)2 as the number of moles of HCl you have because each molecule of Ba(OH)2 has 2 OH groups. So you need .0025 mol of BA(OH)2 to react with the .005 mol of HCl, which, by the same method, also comes out to be .05 L or 50 mL.

hope that helps!
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Answer 2

The simple formula to use is mv = mv

where v is volume and m is molarity

In your first problem it is

(.025)(.200) = X (0.100)

Solve for X and you have liters of NaOH

In the second problem, it is almost the same, except that there are 2 moles of OH for every mole of Ba. Therefore, you have 0.100 moles of OH available.

The answer will be identical to problem #1