Answer to Question #57219 in Physical Chemistry for Kay

Question

How to determine the accurate conc of HCl acid if approximately 3 moldm^-3 HCl acid is provided?
I am suppose to plan an experiment in the lab to answer this question. I am provided with 1.00 moldm-3 NaOH and equipment and materials normally found in a school lab.

Answer 1

Will be necessary for experiment:
1,00 mol/dm3 NaOH solution;
Pipette with a capacity of 5,0 cm³;
The burette for titration with a capacity of 25 cm³.
Measuring cylinder with a capacity of 100 cm³;
Indicator phenolphthalein (alkohol solution);
Flask conic for titration with a capacity of 250 cm3;
The burette to rinse with NaOH solution and to fill it with this solution to a zero mark.
In a flask for titration by means of a pipette to place 5,0 cm³ of the studied HCl solution and by means of the measured cylinder 90 of cm³ of the distilled water (to the total amount of 100 cm³). Then to add 2 – 3 drops of the indicator solution and to titrate with NaOH solution before emergence of the pink coloring steady within 1 minute. The approximate volume of NaOH solution – about 15 cm³.
To calculate exact concentration of solution of hydrochloric acid on a equation:
CM (HCl) = CM (NaOH) * V (NаOH)/V (HCl),
where CM (HCl) - concentration of HCl solution, mol/dm³;
CM (NaOH) - concentration of NaOH solution, mol/dm³;
V (NaOH) – the NaOH solution volume spent for titration, cm³;
V (HCl) - the HCl solution volume taken on titration, cm³.
CM (HCl) = 1,00*V (NaOH)/5,0 = 0,2 * V (NaOH)
It is necessary to repeat experience 2 – 3 times and to calculate mean value.

Answer: CМ(HCl) = 0,2* V(NaOH)