Answer to Question #57219 in Physical Chemistry for Kay
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.
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)
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