Question

If 100 mL of HCl gas at 300 K and 100 kPa dissolved in pure water requires 12.50 mL of the NaOH solution to neutralize in a titration experiment, what is the concentration of the NaOH solution?

Answer 1

The concentration of the NaOH solution is 0.321 mol/L.

1. Write the balanced chemical equation for the reaction.
2. Use the Ideal Gas Law to convert litres of HCl to moles of HCl.
3. Get the molar ratio between HCl and NaOH. This gives you the moles of NaOH.
4. Use the volume of the NaOH to convert moles of NaOH to molarity of NaOH.

Balanced Equation

HCl + NaOH → NaCl + H₂O

Calculate Moles of HCl

`PV = nRT`

`n = (PV)/(RT) = ((100" kPa") ×(0.100" L"))/((8.314" kPa·L·K⁻¹mol⁻¹") ×(300" K"))` = 4.01 × 10⁻³ mol HCl

Calculate Moles of NaOH

4.01 × 10⁻³ mol HCl × `(1" mol NaOH")/(1" mol HCl")` = 4.01 × 10⁻³ mol NaOH

Calculate Molarity of NaOH

Molarity = `"moles of NaOH"/"litres of NaOH soln" = (4.01 × 10⁻³" mol")/(0.012 50" L")` = 0.321 mol/L