A student prepared a base solution using 7 mL of 6M NaOH diluted with distilled water...
- A student prepared a base solution using 7 mL of 6M NaOH diluted with distilled water to a final volume of approximately 400 mL. The student titrated a standardized solution of HCl to determine the exact concentration of the prepared base solution. In the first titration, she found that it took 22.47 mL of the base solution to neutralize 20.00 mL of a 0.98861 M HCl solution. In the second titration, she found that it took 20.12 mL of the base solution to neutralize 17.85 mL of the 0.98861 M HCl solution.
A) What is the molarity of the prepared NaOH solution as determined in titration 1 and titration 2 – calculate them separately.
B) calculate the average molarity of the two trials.
C) How many moles of OH- are present when the endpoint is reached in both trial 1 & trial 2?
*Show your calculations for each trial below.
Homework Answers
The formula that we need to know to solve this problem:-
FOR FIRST TITRATION
Molarity of the standard solution of HCl = 0.98861 M -----(Given)
Volume of the standard solution of HCl taken for titration = 20.00 mL -----(Given)
= (20.00/1000) L
= 0.02 L
No. of moles of HCl present in the standard solution = (0.98861).(0.02) -----using equation(a)
= 0.0197722
= 0.0198 moles
Balanced chemical equation:-
HCl + NaOH ----> NaCl + H2O
It is evident from the balanced chemical equation that for every 1 mole of HCl, 1 mole of NaOH is required for titration.
Hense, for 0.0198 moles of HCl,
No. of moles of NaOH required for titration = 0.0198 moles
Volume of NaOH used in titration = 22.47 mL -----(Given)
= (22.47/1000) L
= 0.02247 L
Molarity of the base solution of NaOH = (0.0198/0.02247) -----using equation(a)
= 0.8811748999
= 0.8811 M
FOR SECOND TITRATION
Molarity of the standard solution of HCl = 0.98861 M -----(Given)
Volume of the standard solution of HCl taken for titration = 17.85 mL -----(Given)
= (17.85/1000) L
= 0.01785 L
No. of moles of HCl present in the standard solution = (0.98861).(0.01785) -----using equation(a)
= 0.0176466885
= 0.0176 moles
Balanced chemical equation:-
HCl + NaOH ----> NaCl + H2O
It is evident from the balanced chemical equation that for every 1 mole of HCl, 1 mole of NaOH is required for titration.
Hense, for 0.0176 moles of HCl,
No. of moles of NaOH required for titration = 0.0176 moles
Volume of NaOH used in titration = 20.12 mL -----(Given)
= (20.12/1000) L
= 0.02012 L
Molarity of the base solution of NaOH = (0.0176/0.02012) -----using equation(a)
= 0.8747514911
= 0.8748 M
Average molarity of the two trials = (Molarity in first titration + Molarity in second titration)/2
= (0.8811 + 0.8748)/2
= 1.7559/2
= 0.87795
= 0.88 M
A) The molarity of the prepared NaOH solution as determined in titration 1 and titration 2 is 0.8811 M and 0.8748 M respectively.
B) The average molarity of the two trials is 0.88 M.
C) No. of moles of OH- present, when the endpoint is reached in both trial 1 and trial 2 is 0.0198 moles and 0.0176 moles respectively.
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