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QUESTION

# How many millimoles of HCl are contained in 5.00 xx 10^1 "mL" of 5.00% (by mass) aqueous HCl solution (density = "1.02 g/mL" )? What volume of 5.00 M NaOH solution would be needed to neutralize 5.00 xx 10^1 "mL" of 5.00% HCl solution?

Here's what I got.

The first thing to do here is to figure out the mass of hydrochloric acid present in that sample of 5.00%"m/m" solution.

You know that the solution has a of "1.02 g mL"^(-1), which means that you get "1.02 g" of solution, i.e. hydrochloric acid and water, for every "1 mL" of solution.

Use the density to find the mass of 5.00 * 10^1"mL" of solution

5.00 * 10^1 color(red)(cancel(color(black)("mL"))) * "1.02 g"/(1color(red)(cancel(color(black)("mL")))) = "51.0 g"

Now, this solution has a 5.00%"m/m" concentration, which means that every "100 g" of solution contain "5.00 g" of hydrochloric acid, the .

This means that your sample will contain

51.0 color(red)(cancel(color(black)("g solution"))) * "5.00 g HCl"/(100color(red)(cancel(color(black)("g solution")))) = "2.55 g HCl"

To convert the mass of hydrochloric acid to moles, use the compound's molar mass

2.55 color(red)(cancel(color(black)("g HCl"))) * "1 mole HCl"/(36.461color(red)(cancel(color(black)("g HCl")))) = "0.06994 moles HCl"

Now, to convert this to millimoles, use the fact that

color(purple)(bar(ul(|color(white)(a/a)color(black)("1 mol" = 10^3"mmol")color(white)(a/a)|)))

You will thus have

0.06994 color(red)(cancel(color(black)("moles HCl"))) * (10^3"mmol")/(1color(red)(cancel(color(black)("mole HCl")))) = color(green)(bar(ul(|color(white)(a/a)color(black)("69.9 mmol HCL")color(white)(a/a)|)))

The answer is rounded to three . Alternatively, you can express this in to get

"no. of mmoles HCl" = 6.99 * 10^1

but I'm not a fan of using scientific notation for values this small.

To answer the second part of the question, use the fact that the of hydrochloric acid with sodium hydroxide

"HCl"_ ((aq)) + "NaOH"_ ((aq)) -> "NaCl"_ ((aq)) + "H"_ 2"O"_ ((l))

consumes equal numbers of moles of the strong acid and the strong base.

This means that in order to neutralize your sample of hydrochloric acid, you must use

"no. of moles of NaOH " = " 69.9 mmoles"

Use the of the solution to find the volume of sodium hydroxide that would contain that many moles -- notice that if you use millimoles you can get the volume in milliliters

69.9 * color(blue)(cancel(color(black)(10^(-3)))) color(red)(cancel(color(black)("moles NaOH"))) * (1 * color(blue)(cancel(color(black)(10^(3))))"mL")/(5.00color(red)(cancel(color(black)("moles NaOH")))) = color(green)(bar(ul(|color(white)(a/a)color(black)("14.0 mL")color(white)(a/a)|)))