Question

SHOW YOUR WORK EXP8 PRELAB ANALYSIS OF ACIDS Follow steps on p. 42 of lab manual PART 1 % ACID IN VINEGAR Molarity of NaOH = 0.092533 M Vinegar volume = 5.00 mL Volume of NaOH = 48.90 mL % Acid in the vinegar = PART 2 % KHP IN IMPURE SAMPLE Molarity of NaOH = 0.092533 M Mass of impure KHP sample used = 1.2294 g Volume of NaOH = 19.39 mL % KHP is sample =

Answer 1

4.11 Solution Part 1: Vinegar is a solution of acetic acid cucoon in water. Now Naou reach with acetic acid in the titration. leiven data is Molarity of Naou = 0.092533 m Vinegar volume = 5.00 mL Naou volume used = 48.90 mL Using Molarity Equation - (Nady) MINI = M₂ V₂ (Vinegar) 0.092533x 48.90 = M2 x 5 - My = 0.092533x48.90 - 0.904972 M E This means molarity of acid in vinegar = 0.904972 m Nowo, 0.904972 M means yooomL of sinepare solution contains acid = 0.904972 moles, . Ime of winegar contains acid = 0.904972 moles. 1000 so. Sme of vinegar contains acid = 0.90497245 -0.004524 moles 1000 I Now Molar Mars of Acetic Acid = 60.052 g/mol. No. of moles can be calculated as I No. of moles = Mans wed in Experiment a Molar man. 4. Experimental Mars = No. of moles & Molar mars = 0.004524 x 60.052 cs Scanned with = 0.2716 CamScanner

% Acid - - Scanned with CamScanner Experimental man x 100 Vinegar Volurne 0.2716 *100 = 5.43% Wlv

Part 2:- Culven data is Molarity of Naou = 0.092533 m Mars of impure' KHP sample wed = 1.22949 Volume of Naou used = 19:39 mL Now molecular formula of KHP is KHCG Hq Ou And Naou reacts with Kup as - KUCG Hq Ou & Naou - KNalg Hyou +4₂0. Now, molarity of Naou is 0.092533 mm This means 1000 ml of solution contains Naou = 0.092533 moles ni Ime of solution contains Naon = 0.092533 moles. 1000 So, 19.39me of solution contains NaOH = 0.092533x19.29 1000 = 0.001994 moles Scanned with CamScanner

o from the chemical HP one mole of the chemical equation, it is clear that NOOH reacts with one mole of KMP -: 0.001794 moles of Naou reach with = 0.001794 moles of KUP. Now , No. of moles is calculated as No.of moles - Experimental man Molar moms. Now, molar man of KUP = 204122 g/mol. .: 0.001194 = Experimental mous 204.22 - Experimental mam = 0·36637 g This means Noon reacts with 0:36637 g of KUP Or we can say Hat 0.36637 9 of Kup is present in impure sample ni Y. KHP im Sample = Mans of Kup xoo Mans of impure sample 0.36637. x 100 = 29.80%. wlw. | Scanned with CamScanner 1.2294