Question

2) 19.63 mL of 0.100 M NaOH is required to neutralize 2.00 mL of a solution containing acetic acid according to the following reaction: HC2H302(aq) + NaOH(aq) → NaC2H3O2(aq) + H2O(1). How many grams of acetic acid are present in 2 mL solution? Report the correct number of significant figures, and report the units. 3) 19.63 mL of 0.100 M NaOH is required to neutralize 2.00 mL of a solution containing acetic acid according to the following reaction: HC2H3O2(aq) + NaOH(aq) → NaC2H3O2(aq) + H2O(l). What is the mass/volume percent of the acetic acid in the solution? Report the correct number of significant figures, and report the units. Mass/Volume Percent of Acetic Acid = Mass of acetic acid (g) Volume of solution (mL) 5 X 100%

Answer 1

2. 19.63 - of mL of 0.100 m Nach, neutroalize a me acetic acid solution - we know, Voronu x nao rash X Swash = Vacètic acid * Sacetic acid re, Viroon = volume of Naoh required = 19. Snach = Stroength of Nooh = 0:1 M Vacetic acid - Volume of acetic acid=2 mL Jacetic acid = strength of "=? us - Vraoy x S NAOH vacetic acid = 19.63 X 0:1 2 = 0.9815 M now, Molecular weight of acetie acid = 60 8m/mole tore, solution 1000 ml loo 0 ) 1 : contains 0.9815 mole 0 .9815 mol X 60 Com mole - 58.89 Som .: 2 me i 1000 = 58.89X2 = 0.11778 gm (-0.118 om fus Ars

NOW - acid mass/volume procent of Acetic = mass of acetic acid (9) - volume of solution (m2)^' -X100% Hre, 0.11778x100 2 = 5.889%. = 5.89% (w As. Hore the actual mars/volume Percent = 7% And the measured value = 5.89% I measured value - actual valid I Percent Emon = - X 100 actual value 15.89-71x100 - 7 3 7 = 15.857% | =15.86%. Imo Ans.