Question

Name ace Section DATA SHEET: EXPERIMENT 16 Date Determination of NaOH Concentration A. Molarity of HCI solution 1 м Trial #1 Trial #2 Trial #3 A6 mt Volume of acid 2 ml. ml. Final buret reading 12 3. ml. mL mL. Initial buret reading 4 mL mL mL 13 Volume of NaOH 5. mL mL ml Molarity of NaOH 6. M М (Show set-up) HCL 25ML0.08mol Hel mol NaoH lmol HCI 0.035L 7. Average molarity of NaOH 155 O

Answer 1

Molarity of NaOH = (Molarity of HCl * Volume of HCl) / Volume of NaOH

Average Molarity of NaOH=(molarity of NaOH in trial#1+molarity of NaOH in trial#2+ molarity of NaOH in trial#3)

3   

13 For Trialt1 Given Holate of Hсе (м) - .08 он volume of the (u) = 25 m2 volume of NaOH (V2) = 13 mL To calculate Molarity of NaoH €M2) = ? calculation - He us NOOH MIVI - M₂ V2 0.08x 95= M₂ X 13 M₂= 0.08x25 Mg = 0.153 M for Teial #2 given - Molarity of Hel (N) = 0.0800 M volume of Hea (v.) = 28 ml volume of NaOH (V2) = 11mL To calculate Molarity of NaOH (M2) = ? Calculation MU vs NOoh MIVI = M₂ V2 0.08x25 = M2 X 11 M₂=0.08x25 MQ = 0.181 M

for Trial #3 Given - of Hel (M) = 0.0800M of re (V) = 25 m2 of NaoH (V2) = 18 ml Molarity Volume Volume To calculate - olarity Calculation - of NaOH (M2) = 8 hce vs Naoh MIVI = MLV₂1 0.08x25= M₂ X 12 Ma = 0.08x25 = THq = 0.166 M 2 Average Molarity of NaOH = Molarity of Naoh in Trial #1 + Molacity of Naoh in Trial 2 olarity of NaOH in Trial #3 + Nola - 0.153 t o.181 t 0.166 0.5 Average Molarity of NaOH = 0.166M And Molarity of NaoH in Trial #1 = 0. 153M Molarity of NaoM in Trial #2 = 0.181 M Molarity of NaoH in Trial #3 = 0.166M Average Molacity of NaOH = 0.166M