Question

CH 111 - Exp 6 Titration of Fruit Juice(2) (2) (Protected View) - Word (Unlicensed Product) Help Tell me what you want to do This experiment is highly similar to the titration of HCI with NaOH so there is no video accompanying this lab. In this titration orange juice is used as the acid in place of HCI. That is because most fruit juices are acidic, with citric acid (H3C6H5O7) being one of the more common acids, especially and not surprisingly in citrus fruits. One big difference between citric acid and HCl is that citric acid is triprotic, meaning it has three protons (H+) to donate, so as you can see in the balanced chemical reaction below, it would take 3 times as many moles of NaOH to neutralize the H3C6H5O7. This makes the molar ratio of H3C6H5O7 NaOH is 13, and it is reflected in the calculations shown below. For this experiment, we assume that all of the acid present in the fruit juice is citric acid to actually make the calculations possible. Na3C6H507(aq) + H2O (aq) Show Calculations H3C6H50+ (aq) + 3 NaOH(aq) — Citric Acid Trial 1 Trial 2 Volume Fruit Juice (mL) 20.0 20.0 Molarity NaOH (m) 0.110 Volume NaOH initial (mL) 0.4 0.6 1 Volume NaOH final (mL) 19.8 20.5 Volume NaOH used in titration (mL) Moles NaOH (mol) Moles H:CeHsOr (mol) Mass H3C6H507 (9) Mass H3C HO7 in 1.0 mL of juice (g/mL) Average Mass in 1.0 mL of juice (g/mL)

Answer 1

H3C6H5O7lag + 3 N90H (aq) 19326H5 07 (aq) + H2O(aq) Trial 1 Thiala volume fruit Juice(ML) 20.0 20.0 molarity MaoH (M) 0.110 volume MaOH initial (m2) 0.4 volume MaoH final (m2) 1978 20.5 in volume naoh used titration (ML) 19.4 то[u hqон Смо) 0.00213 0.00218 moles H3 (64587 (mol) 0.00071 0.00072 mass H3C6H507 (9) 0.136 0.136 0.138 mass H3C6H507 in some of Juice (9lmL) 0-136 0.136 0.138 Average mass in l.ome of Juice (g/m2) 0.137 Scanned with CamScanner

calculation final volume - initial volume volume MaOH used (m2) = for trial 1 H Ma0H used (m2) = used(ML) = volume 19. 8 9.8 - 0.4 = lg.umL for Trial 2 volume MaOH used(ML) = 20.5 0.6 = 19.ame moles Mart - "molarity = moles volume 3 moles = molarity & volume for trial 1 moles maoh = (0-110 ml) (19.4x103) = 0.00213 mol trial 2 moles mqol = 10110 mot) (19.9x10-32) = 0.00218 moll for moles H3C6Hs of a H36H507 lag + 3NaOH(aq) → Naz (6 Hg 07 (aq) + H2Ocagi By mol ratio Trial 1 for (0.00213 mol MaOH) x (1 mol H3C6Hs of) - = (3 mol MaoH) 0.00071 mol H3(6H507 Scanned with CamScanneroles Halo H507 = 0.00071 Mol

for Trial 2 (0.00218 mol) Mach X (1 mol H3(6H507) (3 mol MaoH) = 0.00072 mol H3C6H5o7 moles H316H507 = 0.00072 mol mass H3C6H507(9) in moles mass (g) molal mass (9lmon mass = (moles) (molas mass) molal mass of H3(6H507 = 192.12 9 lmol for trial 1 mass H3 (64507 = (0.00071 mol) (192.12 9 ) = 0.1369 for trial 2 mass H3 (6H507 = 10.00072 moll (192.12.) = 0.1389 Mo mass H3(64507 in 1.0 m2 of Juice (91ML) For Trial 1 mass H3(6H5 oz (9 ) = 0.436 y 1.0m2 = 0r 136 9 Ime for Trial 2 mass H3(6h5o7 (91= 9.138 g = 0.138 91m2 Average mass in 1.0m2 of Juice (9lmL) ra Scanned notin 36 g/mL + 0.138 gimL CamScanner scs 0.137 g/m2