REPORT SHEET LAB 20 Acid-Base Titration A. Concentration of...

REPORT SHEET LAB 20 Acid-Base Titration A. Concentration of Acetic Acid in Vinegar 1. Brand Lady's Choice Volume 5.0 mL (% on label) 4 % 2. Molarity (M) of NaOH 0.10 M Trial 1 Trial 2 Trial 3 3. Initial NaOH level in buret 0.22 mL 1.17 mL 2.43 mL 4. Final NaOH level in buret 33.62 mL 33.49 mL 48.37 mL 5. Volume (mL) of NaOH used 33.40 mL 33.32 mL 51.82 mL 6. Average volume (mL) 39.18 mL 7. Average volume in liters (L) 0.03918 L 8. Moles of NaOH used in titration mole NaOH (Show calculations.) 9. Moles of HC2H3O2 neutralized by NaOH mole HC2H3O2 10. Molarity of HC2H3O2 M HC2H3O2 (Show calculations.) 11. Grams of HC2H3O2 g HC2H3O2 (Show calculations.) 12. Percent (m/v) HC2H3O2 in vinegar % HC2H3O2 (Show calculations.)

Transcript

00:01 Hello everyone in this question we have been given a lab report of an acid base titration and we are asked some questions for which we have to give the answer.

00:17 So in the eighth question we are asked the number of moles of noh.

00:26 So in order to find out the number of moles of naoh we are going to use the formula that is molarity number of moles is equal to molarity into volume so here the molarity that is given to us is 0 .1 and the volume that was found out was 0 .03918 liters so if we multiply this we will get 0 .003 918 moles and if we find if we write it in a better way it can be written as 3 .918 into 10 raised to the power minus 3 moles so this will be the answer of the first question then in the ninth part we are said to find out the moles of a c h c2 h3 o2 which is neutralized by neutralized by an a which for that we are going to see the reaction that is actually taking place which is naoh in aqua solution reacts with ch3 cooh in aqua solution to give ch3 c ona plus h2o so what we are going to do is we are going to divide the number of of moles of mh3 cooh with its coefficient with is going to be equal to the number of moles of n -a -o -h divided by its coefficient.

02:36 Okay, so the coefficient on both the side is 1 so that will get cancelled out and we know that the number of moles of n -a -o -h is this so the number of moles of the number of moles of ch3 cooh becomes equal to the number of moles of naoh which we already found out which is 3 .918 into 10 raised to the power minus 3 moles.

03:08 So this will be the answer.

03:11 Then on the 10th part we are asked to find out the molarity of molarity of hc2 h3 o2 so for that we are going to use the formula moles divided by number of moles divided by volume in liters so this is the formula for finding out the molarity so if we calculate this we know the number of moles that is 3 .918 into 10 rish to the power minus 3 and the volume in liters is 5 into 10 rous to the power minus 3.

03:52 So that will come out to be 0 .7836 molar.

03:58 Then on the 11th part we are asked to find out the mass of hc2h3o2...

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