Question

Quantitative Analysis of Vinegar via Titration

Lab Team Members Quantitative Analysis of Vinegar via Titration Report Sheet 1 Part A: Standardizing a Dilute Sodium Hydroxide, NaOH(aq). Solution KHC,H,O,(aq) + NaOH(aq) + H2O(1) + KNaC,H,O,(aq) 204.22 g/mol Trial 1 Trial 2 Trial 3 10.33919 0.36769 0.35989 moles of KHP, mol moles of NaOH, mol initial buret reading, ml 4.00 ml 20.10 mi 30.10 m final buret reading, ml 19.30 ml 35.60m145.10 ml volume of NaOH, ml. volume of NaOH, LL [NaOH], M average (NaOH), M Show your calculations for Trial I on the back of this page.

Answer 1

Mass Calculate the moles of" KHPbyu sin gthe formula Moles=; -as follows: Molar mass Trial 1molesof KHP- Trallmass of KHP 0.33919KHP -=0.0016609 mol KHP Molar mass of KHP 204.22g/mol Trial 2 moles of KHP= Trial2 mass of KHP 0.36769g KHP 22=0.0081005 mol KHP Molar mass of KHP 204.22g/mol Trial3 mass of KHP 0.36769g KHP Trial 3moles of KHP= -=0.0071623 mol KHP Molar mass of KHP 204.22g/mol -------------------------------------------------- From the Balanced reaction thatis KHCH,0(aq)+NaOH(aq)— ,0()+KNaC,H,9 1mole of KHC 1.0 reacts withone mole of NaOH. Thus, moles of Nao Has follows: Trial Imoles of NaOH = 0.0016609 mol KHP Imol NaOHO -=0.0016609 vol NaOH Imol KHP=" Trial 2 moles of NaOH = 0.0081005 mol KHPX - Imol NaOH 0. 0081005aal NaOH 1mol KHP Trial 3moles of NaOH = 0.0071623 mol KHPX- -=0.0071623mol NaOH 1 mol KHP ------------------------------------------------------ Volume of NaOH = Final volume of buret - Initial volume of buret Calculate the volume of NaOH as follows: Trial 1Volume of NaOH = Final volume of buret - Initial volume of buret =19.30mL-4.90 mL =14.4 mL of NaOH Trial 2 Volume of NaOH = Final volume of buret - Initi al volume of buret = 35.66mL-20.10 mL =15.56 mL of NaOH Trial 3Volume of NaOH = Final volume of buret - Initial volume of buret = 45.10mL-30.10mL =15.00 mL of NaOH ---------------------------------------------

To convert volume of mL into Luse the conversion factor that is in 12 1000 m2 Calculate the volume of NaOH in Las follows: Trial 1Volume of NaOH =14.4mL of NaOH 12 --_=1.44 x 10-22 1000 ml 12 Trial 2 Volume of NaOH =15.56 mL of NaOH X- 1000 mL =1.556 x 10-22 Trial 3Volume of NaOH = 15.50 mL of NaOH * 12 -=1.550x10-22 1000 ml - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - Moles of NaOH Calculate the concentration of NaOH byu sing formula concentration=- Volume of NaOH inL Calculate the concentration of NaOH as follows: 0.0016609mol NaOH Trial 1Concentationof NaOH => E=0.1153 M NaOH 1.44x1024 . 0.0081005a: NaOH Trial 2 Concentationof NaOH = ==0.52056M NAOH 1.556 x 10-²2 0.0071623mol NaOH Trial 3Volume of NaOH = -=0.4621M MOH 1.550x10- - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - Calculate the average of[NaOH]Mas follows: 0.1153 M NaOH +0.5206M NAOH +0.4621M NaOH --=0.3660 M Updated tableisas follows: Trial 1 Trial 2 Trial 3 | Mass of KHP 0.33919g | 0.36769g 0 .36769g Molesof KHP 0.0016609mol 0.0081005 mol 0.0071623mol Moles of NaOH 0.0016609 mol 0.0081005 mol 0.0071623mol Initial buret volume | 4.90ml | 20.10ml 30.10 mL Final buret volume 19.30 ml 35.66 m2 45.10mL Volume of NaOH inml 14.4 m2 15.56 m2 15.50 ml Volume of NaOH in L | 1.44 x 10-22 | 1.556 X10-L 1.550X10-2 L [MaOH]M 0.1153 M 0.52056M 0.4621M Average [NaOH]M 0.3660M

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