Learners perform a titration to standardise a dilute hydrochloric acid solution.
They use 25 cm3
of a standard 0,15 mol∙dm–3 sodium hydroxide (NaOH) solution.
The titration is repeated three times, the average HCl volume readings are calculated and found
to be 30 cm3
.
3.1 Calculate the pH of the NaOH solution?
The balanced reaction for the above titration is:
NaOH(aq) + HCl(aq) → NaCl(aq) + H2
O(l)
Calculate the concentration of the HCl.?
To calculate the pH of the NaOH solution, we need to determine the concentration of hydroxide ions (OH-) in the solution.
Given:
Volume of NaOH solution used = 25 cm3
Molarity of NaOH solution = 0.15 mol/dm3
Using the equation for the reaction:
NaOH(aq) + HCl(aq) → NaCl(aq) + H2O(l)
The reaction is a 1:1 molar ratio, so the number of moles of NaOH used is equal to the number of moles of HCl.
Number of moles of NaOH used = Molarity × Volume
= 0.15 mol/dm3 × 0.025 dm3
= 0.00375 mol
Since the reaction is 1:1, the number of moles of HCl used is also 0.00375 mol.
The volume of HCl used (average reading) is 30 cm3, which is equal to 0.03 dm3.
Concentration of HCl = Number of moles / Volume
= 0.00375 mol / 0.03 dm3
= 0.125 mol/dm3
Therefore, the concentration of the HCl solution is 0.125 mol/dm3.