Question

In a titration, 50.00 cm3 of 0.300 mol/dm3

sodium hydroxide solution is exactly neutralized by 25.0
cm3 of a dilute solution of hydrochloric acid.
NaOH(aq) + HCl(aq) → NaCl(aq) + H2O(l)
Calculate the concentration of the hydrochloric acid. Show your work out

PLS URGENT ANS

Answer 1

Step 1: Calculate the amount of sodium hydroxide in moles

Volume of sodium hydroxide solution = 50.0 ÷ 1,000 = 0.05 dm3

Rearrange:

Concentration in mol/dm3 = amount of solute in molvolume in dm3

Amount of solutein mol = concentration in mol/dm3 × volume in dm3

Amount of sodium hydroxide = 0.300 × 0.05

= 0.01 5mol

Step 2: Find the amount of hydrochloric acid in moles

The balanced equation is: NaOH(aq) + HCl(aq) → NaCl(aq) + H2O(l)

So the mole ratio NaOH:HCl is 1:1

Therefore 0.015 mol of NaOH reacts with 0.015 mol of HCl

Step 3: Calculate the concentration of hydrochloric acid in mol/dm3

Volume of hydrochloric acid = 25.00 ÷ 1000 = 0.025 dm3

Concentration in mol/dm3 = amount of solute in molvolume in dm3

Concentration in mol/dm3 = 0.015/0.025

= 0.6 mol/dm3

Step 4: Calculate the concentration of hydrochloric acid in g/dm3

Relative formula mass of HCl = 1 + 35.5 = 36.5

Mass = relative formula mass × amount

Mass of HCl = 36.5 × 0.6

= 21.9 g

So concentration = 21.9 g/dm3