Question

. A series of solutions containing NaOH, Na2CO3, and NaHCO3, alone or in compatible combination, was titrated with 0.1202 M HCl. Tabulated below are the volumes of acid needed to titrate 25.00-ml portions of each solution to (1) a phenolphthalein and (2) a bromocresol green end point. Use this information to deduce the composition of the solutions. In addition, calculate the mass in milligrams of each solute per milliliter of solution. (1) (2) (a) 22.42 22.44 (b) 15.67 42.13 (c) 29.64 36.42 (d) 16.12 32.23 (e) 0.00 33.333

PLEASE ANSWER ALL QUESTIONS (A-E, PLEASE SHOW STEP BY STEP CALCULATION (NOT THE EXCEL SHEET)

THIS IS AN ANALYTICAL CHM QUESTION

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Answer 1

for letr I for Naon, Case I As we know Mix, = M₂ V ₂ (for 1st Hading) (HU) (Nady with kembahtle 0:1 202 X 22.42 = M2 - M2 = 0.1079 Case I. Mivi - M₂ V2. (Ind reading M₂ = 0.1202 x 15.67 - - » M, = 0.075 ② M₂ Similarly, M2 = 0.1202 % 29.64 = 0.142 - M3 21" Case IIT. urd reading Cose to similarly, M2 = 0.1202 X 16.12 = 0.0005 4 My Case I Now M₂ = 0.1202 x 0. 0 6 . Mr. for total molarity , we can use the formula, M = MIN + M2 V, + M₂ V3 + MyVy + M585 - M, + M2 + M₂ + My + M5 = 10.107*25) +(0.075 *25) + (0.1422 25) +(0.0778 X 25) + 0 0.107 +0.075 +0.142 +0.0775 M = 10.037 0.4015 - 25ML How from the formula, M = no. of moles of solute (no) Volume of soln (V) nA = MAV = 25mx 0.025ml = 0.625 "HB - nox MB Li Weight of solute for Neon = 0.625x40 Two = 25 mg

17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 for Na, con no. of moles = weight of each solute I noleular weight colulé 0.625 = weight of each loc da 66.24 mg & weight of Guen in question volume is Na, Ozan at mal. For Natco, 0.625 = weight of Nanco 52.50 mg V 084.01 ) o weight of Natico in the solution