18. A sodium hydroxide solution (25.0 cm3 ) is neutralized by 16.0 cm3 of a solution of
sulfuric acid of concentration 0.01 mol dm-3. Find the NaOH concentration.
19. Calculate the concentration (in mol dm-3 ) of 3% (percentage by mass) solution of
potassium iodide. The density of the solution is assumed to be 1.
Relative atomic masses: K = 39, I = 127.
Answers
Answer:
Worked example
In a titration, 25.0 cm3 of 0.100 mol/dm3 sodium hydroxide solution is exactly neutralised by 20.00 cm3 of a dilute solution of hydrochloric acid. Calculate the concentration of the hydrochloric acid solution.
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 = \frac{\textup{amount~of~solute~in~mol}}{\textup{volume~in~dm}^3}
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 = \frac{\textup{amount~of~solute~in~mol}}{\textup{volume~in~dm}^3}
Concentration in mol/dm3 = \frac{\textup{0.00250}}{\textup{0.0200}}
= 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
Explanation:
In a titration, 25.0 cm3 of 0.100 mol/dm3 sodium hydroxide solution is exactly neutralised by 20.00 cm3 of a dilute solution of hydrochloric acid. Calculate the concentration of the hydrochloric acid solution.
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 = \frac{\textup{amount~of~solute~in~mol}}{\textup{volume~in~dm}^3}
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 = \frac{\textup{amount~of~solute~in~mol}}{\textup{volume~in~dm}^3}
Concentration in mol/dm3 = \frac{\textup{0.00250}}{\textup{0.0200}}
= 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