Question #636b5

1 Answer
May 23, 2016

Answer:

It will take 9.0 mL of this acid to neutralize 1 L of 0.1 mol/L #"NaOH"#.

Explanation:

There are three parts to this problem:

  1. Calculate the moles of #"HCl"#.
  2. Calculate the molarity of the #"HCl"#.
  3. Do the titration calculation.

Moles of HCl:

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

#color(blue)(|bar(ul(color(white)(a/a)PV = nRTcolor(white)(a/a)|)))" "#

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

#"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"#

#"Mass of solution" = "1 g + 0.5232 g" = "1.532 g"#

#"Volume of solution" = 1.532 color(red)(cancel(color(black)("g"))) × "1 mL"/(1.18 color(red)(cancel(color(black)("g")))) = "1.291 mL"#

#"Molarity" = "moles"/"litres" = "0.014 35 mol"/"0.001 291 L" = "11.1 mol/L"#

The titration calculation

The equation for the reaction is

#"HCl + NaOH" → "NaCl" + "H"_2"O"#

#"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"#

#"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)