Question

can someone please help me, this is a dry lab and im so lost ..

Data Table Part I: Standardization of NaOH Trial One Trial Two 2.1245 3.2019 1.) Mass of empty weigh boat (9) 2.) Mass of weigh boat + solid KHP (9) 3.) Mass of solid KHP (9) Show calculation for Trial One: 8.56 19.85 4.) Initial volume of NaOH (mL) 5.) Final volume of NaOH (mL) 6.) Volume of NaOH used (ml) Show calculation for Trial One! 7.) Volume of NaOH used (L) Show calculation for Trial One: 8.) Moles of KHP (mols) Show calculation for Trial One: 9.) Moles of NaOH (mols) Show calculation for Trial One: 10.) Molarity of NaOH (M) Show calculation for Trial One: 11.) Average Molarity of NaOH (M) Show calculation:

Answer 1

for calculation of part (II), Volume of Vinegar is needed.

Part I : 3 Mass of solid KHP (g) = Mass of weight Boat +KHP) -.- Mass of weight Boat) = 3.20 19 – 2.1245 = 1,0774 to Volume of NaOH used (mL) = final volume - Initial Volo = 19.85-8.56= 11-29 I Volume of NaOH used (2) = Volume of Na OH used (m2) _love - 11-29 = 0.01129: As 12 = 1000 me) 1000 & Males of KHP used = Mass of KHP (mols Molar Mass of KHP = 1.0774 - 0.0052 757 204.22 KHP+ NaOH → Nakp + H2O 9 As 1 mol KHP requires 1 mol NaOH - to React completely. Lo moles of NaOH = moly of K HP = 0.0052757 (mols) 10 Molwrity of NaOH (M) = No of moles - Volume of NaOH (2) = 0.0052757 - 0.4673 - 0.01129 11 As there is data only for 1 Toial C Average Molarity of Na OH (M = 0.4673