# Question 636b5

May 23, 2016

It will take 9.0 mL of this acid to neutralize 1 L of 0.1 mol/L $\text{NaOH}$.

#### Explanation:

There are three parts to this problem:

1. Calculate the moles of $\text{HCl}$.
2. Calculate the molarity of the $\text{HCl}$.
3. Do the titration calculation.

Moles of HCl:

We can use the Ideal Gas Law to calculate the moles of gas.

$\textcolor{b l u e}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} P V = n R T \textcolor{w h i t e}{\frac{a}{a}} |}}} \text{ }$

This gives

n = (PV)/(RT) = (101.325 color(red)(cancel(color(black)("kPa"))) × 0.3452 color(red)(cancel(color(black)("L"))))/(8.314 color(red)(cancel(color(black)("kPa·L·K"^"-1")))"mol"^"-1" × 293.15 color(red)(cancel(color(black)("K")))) = "0.01 435 mol"#

Molarity of the HCl

$\text{Mass of HCl" = "0.014 35" color(red)(cancel(color(black)("mol HCl"))) × "36.46 g HCl"/(1 color(red)(cancel(color(black)("mol HCl")))) = "0.5232 g HCl}$

$\text{Mass of solution" = "1 g + 0.5232 g" = "1.532 g}$

$\text{Volume of solution" = 1.532 color(red)(cancel(color(black)("g"))) × "1 mL"/(1.18 color(red)(cancel(color(black)("g")))) = "1.291 mL}$

$\text{Molarity" = "moles"/"litres" = "0.014 35 mol"/"0.001 291 L" = "11.1 mol/L}$

The titration calculation

The equation for the reaction is

$\text{HCl + NaOH" → "NaCl" + "H"_2"O}$

$\text{Moles of HCl" = 1 color(red)(cancel(color(black)("L NaOH"))) × (0.1 color(red)(cancel(color(black)("mol NaOH"))))/(1 color(red)(cancel(color(black)("L Na OH")))) × "1 mol HCl"/(1 color(red)(cancel(color(black)("mol NaOH")))) = "0.1 mol HCl}$

$\text{Volume of HCl" = 0.1 color(red)(cancel(color(black)("mol HCl"))) × "1 L HCl"/(11.1 color(red)(cancel(color(black)("mol HCl")))) = "0.0090 L" = "9.0 mL}$
(2 significant figures)