# Question 2985b

Sep 30, 2016

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 ${\text{1.02 g mL}}^{- 1}$, which means that you get $\text{1.02 g}$ of solution, i.e. hydrochloric acid and water, for every $\text{1 mL}$ of solution.

Use the density to find the mass of $5.00 \cdot {10}^{1} \text{mL}$ 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 $\text{100 g}$ of solution contain $\text{5.00 g}$ 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

$\textcolor{p u r p \le}{\overline{\underline{| \textcolor{w h i t e}{\frac{a}{a}} \textcolor{b l a c k}{\text{1 mol" = 10^3"mmol}} \textcolor{w h i t e}{\frac{a}{a}} |}}}$

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

$\text{no. of mmoles HCl} = 6.99 \cdot {10}^{1}$

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

${\text{HCl"_ ((aq)) + "NaOH"_ ((aq)) -> "NaCl"_ ((aq)) + "H"_ 2"O}}_{\left(l\right)}$

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

$\text{no. of moles of NaOH " = " 69.9 mmoles}$

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)|)))#