Question

If 100 mL of HCl gas at 300 K and 200 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 0.641 mol/L.

There are four steps to this calculation.

1. Write the balanced chemical equation for the reaction.
2. Calculate the moles of HCl.
3. Convert moles of HCl → moles of NaOH.
4. Calculate the molarity of the NaOH.

Step 1.

HCl + NaOH → NaCl + H₂O

Step 2.

`PV = nRT`

`n = (PV)/(RT) = (200"kPa" × 0.100"L")/(8.314"kPa·L·K⁻¹mol⁻¹" × 300"K")` = 8.019 × 10⁻³ mol

Step 3.

Moles of NaOH = 8.019 × 10⁻³ mol HCl × `(1"mol NaOH")/(1"mol HCl")` = 8.019 × 10⁻³ mol NaOH

Step 4.

Molarity of NaOH = `"moles of NaOH"/"litres of solution" = (8.019 × 10⁻³"mol")/(0.01250"L")` = 0.641 mol/L (3 significant figures)

The molarity of the NaOH is 0.641 mol/L.