Question

1. 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. 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. What is the molarity of the prepared NaOH solution as determined in titration 1 and titration 2 - calculate them separately. Afterwards, calculate the average molarity of the two trials. How many moles of OH are present when the end point is reached in both trial 1 & trial 2? *Show your calculations for each trial below. What color should the solution appear at the end point?

Answer 1

As per the Formula N1V1=N2NV2

For first titration - The Molarity of NaOH is N1 , which is -

N1 * 22.47 = 20 * 0.98861

N1 = 19.7722 / 22.47

N1 = 0.879937 M

So the conentraion of prepared NaOH is 0.8799 M

For second titration, similarly

N1 * 20.12 = 17.85 * 0.98861

N1 = 17.646688 / 20.12

N1 = 0.87707 M

so the concentration of NaOH in second titration is 0.87707 M

Average Molarity of two trials = (0.8799 + 0.87707) / 2 = 0.8753 M

For the reaction , remailning moles of OH ions calculations is shown in attached sheet.

please find it.

At the end point if the indicator used is Phenolphthalein indicator, it will give a pink color as the end point and if Methyl orange is used as an indicator it will show first pink color then at the end point it becomes colorless.

Nace & H₂O i NaOH + He - 0.0753 0.98661 M As per the formula W M = 7 - no of moles volume (L) for trial- NaOH- .8753 = n= 0.019667 moles 7 22.47 x103 3 ) n = 0.01761 moles 20.12 x 10 trial - 2 - 0.8753 = for Hee - trial- - 0.98961 = n n = 0.01977 moles trial-2- 0.9886) = 17.85 x 163 n= 0.01764 moles It is clearly in bath trials free amount seen after of from the OH the above calculations that end point, there is no ions.