To a #0.1# #M# acetic acid solution with volume of #100# #ml#, #0.1# #M# #NaOH# solution is added. Now, the acetic acid concentration becomes #0.05# #M#. What are the number of molecules present in #100# #mu##L# of #0.1# #M# #NaOH# solution?

I'm assuming that there's more to the problem than what you have here, but either way all you have to do is use the molarity and volume of the sodium hydroxide solution to determine how many moles of solute it contains.

Once you know that, use Avogadro's number to find the number of moles of sodium hydroxide.

Now, the volume is given to you in microliters, #mu"L"#, so make sure that you convert it to liters by using the conversion factor

#color(purple)(|bar(ul(color(white)(a/a)color(black)("1 L" = 10^6mu"L")color(white)(a/a)|)))#

So, a #"0.1 M"# solution of sodium hydroxide contains #0.1# moles of solute per liter of solution. This means that #100mu"L"# will contain

#100 color(red)(cancel(color(black)(mu"L"))) * (1color(red)(cancel(color(black)("L"))))/(10^6color(red)(cancel(color(black)(mu"L")))) * "0.1 moles NaOH"/(1color(red)(cancel(color(black)("L")))) = 10^(-4)"moles NaOH"#

As you know, one mole of any substance contains exactly #6.022 * 10^(23)# molecules of that substance -- this is known as Avogadro's number.

In your case, sodium hydroxide is an ionic compound, so you don't really have molecules of sodium hydroxide, you have formula units.

#color(blue)(|bar(ul(color(white)(a/a)"1 mole" = 6.022 * 10^(23)"f. units"color(white)(a/a)|)))#

You will thus have

#10^(-4)color(red)(cancel(color(black)("moles NaOH"))) * (6.022 * 10^(23)"f. units")/(1color(red)(cancel(color(black)("mole NaOH")))) = color(green)(|bar(ul(color(white)(a/a)color(black)(6 * 10^(19)color(white)(a)"f. units")color(white)(a/a)|)))#

The answer is rounded to one sig fig.

SIDE NOTE An important thing to point out here is that technically your sodium hydroxide solution cannot have a molarity of #"0.1 M"#.

Sodium hydroxide dissociates completely in aqueous solution to form sodium cations, #"Na"^(+)#, and hydroxide anions, #"OH"^(-)#.

#"NaOH"_ ((aq)) -> "Na"_ ((aq))^(+) + "OH"_ ((aq))^(-)#

An accurate description of a #"0.1 M"# sodium hydroxide solution would be a solution that is #"0.1 M"# in #"Na"^(+)# and #"0.1 M"# in #"OH"^(-)#.

Now, you can also use formality instead of molarity to describe the solution

Formality doesn't take into account what happens to the solute once you dissolve it in solution.

So even if you're dealing with a strong electrolyte that dissociates completely in aqueous solution, you can express the solution's concentration without worrying about what happens to the solute in solution.

This means that you can express the solution as #"0.1 F NaOH"#, or #"0.1 formal NaOH"#.

So, to sum this up, your solution is

#{("0.1 M Na"^(+)), ("0.1 M OH"^(-)):} " " " " "OR" " " " " "0.1 F NaOH"#

Formality is much less common than molarity, which is why you'll often see things like #"0.5 M NaCl"# or #"1.0 M HCl"#.

In these cases, it is understood that the solute dissociates in aqueous solution, so molarity is still used instead of formality.