Question

Question #2985b

Answer 1

Here's what I got.

Explanation:

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 density 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 solute.

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 sig figs. Alternatively, you can express this in scientific notation 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 neutralization 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 molarity 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)|)))`