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
.
- Calculate the pH of the NaOH solution
To calculate the pH of the NaOH solution, we need to find the concentration of hydroxide ions (OH-) in the solution.
First, let's calculate the number of moles of NaOH used in the titration:
Moles of NaOH = Concentration (mol/dm^3) x Volume (dm^3)
= 0.15 mol/dm^3 x 0.025 dm^3
= 0.00375 mol
Since NaOH is a strong base, it fully dissociates in water to produce one hydroxide ion for every one NaOH molecule. So, the number of moles of OH- ions in the solution is also 0.00375 mol.
Next, let's calculate the total volume of the NaOH solution used in the titration:
Total Volume = Volume used in titration x Number of times the titration was repeated
= 0.03 dm^3 x 3
= 0.09 dm^3
Now, we can calculate the concentration of OH- ions in the NaOH solution:
Concentration (mol/dm^3) = Moles of OH- ions / Total Volume (dm^3)
= 0.00375 mol / 0.09 dm^3
= 0.0417 mol/dm^3
Since the concentration of OH- ions is equal to the concentration of H+ ions in a neutral solution, the concentration of H+ ions in the NaOH solution is also 0.0417 mol/dm^3.
Finally, let's calculate the pH of the NaOH solution:
pOH = -log10(OH- concentration)
= -log10(0.0417)
= 1.38
pH = 14 - pOH
= 14 - 1.38
= 12.62
Therefore, the pH of the NaOH solution is 12.62.