A student prepared a base solution using 7 mL of 6M NaOH diluted with distilled
ID: 1031744 • Letter: A
Question
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 HCI to determine the exact concentration of the prepared base solution. She found that it took 24.55 mL of the base solution to neutralize 20.00 mL of a 0.99897 M HCI solution 1. What is the molarity of the prepared NaOH solution as determined in the titration? How many moles of OH are present when the end point is reached? What color should the solution appear at the end point?Explanation / Answer
The reaction is given as :
HCl + NaOH = NaCl + H2O
So equimolar amounts of both the reagents are used.
So using the formula M1v1 = M2V2
where M is molarity and v is volume
0.99897 x 20.00 = M x 24.55
M = 0.81382 M
So the molarity of NaOH solution will be 0.81382 M
Since at the end point , all of the acid is neutralised with the base , so the moles of OH- present will be 0.
The pH indicator phenolphthalein is used to detect the end point. So pink color appears in the solution when the end point is reached.
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