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?

Question

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?

Olivia Anton · Accepted Answer

Here's what I got.

In any case, all you need to do is calculate how many moles of solute the sodium hydroxide solution contains using its volume and molarity. I'm assuming there's more to the problem than what you have here. Once you have that information, calculate the moles of sodium hydroxide using Avogadro's number. 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. Since sodium hydroxide is an ionic compound in your situation, you actually have formula units rather than molecules of sodium hydroxide. #color(blue)(|bar(ul(color(white)(a/a)"1 mole" = 6.022 * 10^(23)"f. units"color(white)(a/a)|)))# Consequently, you will 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)|)))# One sig fig is used to round the result. 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"^(-)#. Currently, formality can also be used to describe the solution in place of molarity. Formality disregards the behavior of the solute after it dissolves in solution. Therefore, you can express the concentration of the solution without worrying about what happens to the solute in solution, even if you're working with a strong electrolyte that dissociates completely in aqueous solution. This means that you can express the solution as #"0.1 F NaOH"#, or #"0.1 formal NaOH"#. In summary, therefore, your answer 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"#. Since it is known that the solute dissociates in aqueous solution in these situations, formality is still avoided in favor of molarity.

Owen Ambrose · Answer

undefined The number of molecules present in 100 μL of 0.1 M NaOH solution is approximately $ 6.022 \times 10^{17} $ molecules.