There are three parts to this problem:
- Calculate the moles of #"HCl"#.
- Calculate the molarity of the #"HCl"#.
- 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)
