chemistry help!!?

Question

can someone please help me answer these?
i have no idea how to =(.

if 110.0mL of 3.00 M sulfuric acid has 25.0 mL of water added to it. What is the resulting concentration of the solution.

how much water must be added to a 50.0 mL sample of 18.0 M sulfuric acid to gie a resulting concentration of 0.250 M?

What is the sodium ion concentration when 250.0 mL of water is added to 125.5 mL of a 3.21 M solutio of sodium phosphate?

i tried finding the instructions to these kinds of questions in my text book.. but i can't find anything =(

thanks in advance !

Answer 1

You might want to look again; this is pretty basic stuff.
FIRST RULE: In dilution problems, the original number of moles of the solute dont change. The volume of the solvent does. In short
moles = [Molar conc] x [volume in L]

In problem 1= How much sulfuric acid do you have in MOLES? The answer is .11 L x 3 Molar= 0.33 moles. Now we add 25 ml of water. We have 0.135 L, but still 0.33 moles. So the answer will be 0.33 moles/0.135 L as Molar.

In problem 2, we ask again. How much sulfuric acid do you have in MOLES? The answer is 0.05x 18 M = 0.9 moles. Now we add water until the conc. is 0.250 mole/L. At that point, we have a total volume of 0.9 moles/0.250 moles/L. We started with 0.05 L and added the rest.

In problem 3, we have to know that for each molecule of sodium phosphate, we have one ion of sodium in solution; the sodium completely ionizes. So we start with how much sodium ion?
That's right, 0.1255 L x 3.21 moles/Liter . At the end, we still have that much sodium ion, but the volume is 0.3755 L. You should be able to figure the rest.