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# 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)|)))##