Imagine you had HCl with a concentration of exactly 0.10 mol/dm3. If 0.023 dm3 of a sodium hydroxide solution, NaOH (aq), could exactly neutralize 0.040 dm3 of the HCl solution, what is the concentration of the NaOH (aq)
Question
Question
Imagine you had HCl with a concentration of exactly 0.10 mol/dm3. If 0.023 dm3 of a sodium hydroxide solution, NaOH (aq), could exactly neutralize 0.040 dm3 of the HCl solution, what is the concentration of the NaOH (aq)
Answer:
Explanation:
Step 1: Calculate the amount of sodium hydroxide in moles
Volume of sodium hydroxide solution = 25.0 ÷ 1,000 = 0.0250 dm3
Rearrange:
Concentration in mol/dm3 =
Amount of solutein mol = concentration in mol/dm3 × volume in dm3
Amount of sodium hydroxide = 0.100 × 0.0250
= 0.00250 mol
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 0.00250 mol of NaOH reacts with 0.00250 mol of HCl
Step 3: Calculate the concentration of hydrochloric acid in mol/dm3
Volume of hydrochloric acid = 20.00 ÷ 1000 = 0.0200 dm3
Concentration in mol/dm3 =
Concentration in mol/dm3 =
= 0.125 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 × 0.125
= 4.56 g
So concentration = 4.56 g/dm3