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?

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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?

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Answer 1 · 2014-08-10T19:00:37

The concentration of the NaOH solution is 0.641 mol/L.

There are four steps to this calculation.

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

Step 1.

HCl + NaOH → NaCl + H₂O

Step 2.

$P V = n R T$ $PV = nRT$

$n = \frac{P V}{R T} = \frac{200 kPa \times 0.100 L}{8.314 kPa\cdotL\cdotK⁻¹mol⁻¹ \times 300 K}$ $n = (PV)/(RT) = (200"kPa" × 0.100"L")/(8.314"kPa·L·K⁻¹mol⁻¹" × 300"K")$ = 8.019 × 10⁻³ mol

Step 3.

Moles of NaOH = 8.019 × 10⁻³ mol HCl × $\frac{1 mol NaOH}{1 mol HCl}$ $(1"mol NaOH")/(1"mol HCl")$ = 8.019 × 10⁻³ mol NaOH

Step 4.

Molarity of NaOH = $\frac{moles of NaOH}{litres of solution} = \frac{8.019 \times 10 ⁻ ³ mol}{0.01250 L}$ $"moles of NaOH"/"litres of solution" = (8.019 × 10⁻³"mol")/(0.01250"L")$ = 0.641 mol/L (3 significant figures)

The molarity of the NaOH is 0.641 mol/L.

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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?

1 Answer

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