Learner perform a titration to standardise a dilute hydrochloric acid solution. They use 25cm^3 of a standard 0,15 mol.dm^-3 sodium hydroxide (NaOH) solution. the titration is repeated three times, the average HCI volume readings are calculated and found to be 30 cm63
1. calculated the pH of the NaOH solution
2. give a reason why the titration is repeated more than once
3. the balanced reaction for the above titration is
NaOH(aq)+HCI(aq)->NaCI(aq)+H2O(I)
calculate the concentration of the HCI
1. To calculate the pH of the NaOH solution, you would need to know the concentration of the NaOH solution. However, the concentration is not provided in the given information. Without the concentration, it is not possible to calculate the pH.
2. The titration is repeated more than once to obtain more accurate results and to reduce any potential errors or discrepancies in the measurements. A single titration may not always provide precise and reliable results, so repeating the titration multiple times helps to minimize errors and obtain a more accurate average value.
3. The balanced reaction for the titration is:
NaOH(aq) + HCl(aq) -> NaCl(aq) + H2O(l)
To calculate the concentration of HCl, you would need to use the balanced equation and the known information about the NaOH solution. The mole ratio between NaOH and HCl in the balanced equation is 1:1.
Given that 25 cm^3 of the NaOH solution with a concentration of 0.15 mol.dm^-3 is required to neutralize the HCl solution, we can use the following equation to calculate the concentration of HCl:
0.15 mol.dm^-3 NaOH x 0.025 dm^3 = x mol.dm^-3 HCl x 0.025 dm^3
Simplifying the equation:
0.00375 mol = x mol
Therefore, the concentration of the HCl solution is 0.00375 mol.dm^-3.