Question

I need to answer 1 , 2 molarity with steps and answer 3 , 4.

Name: CHE 121 Lab: Titration Purpose: To determine the M of an unknown NaOH solution and then to determine the M of vinegar. Reaction 1: HCI + NaOH → NaCl + H2O Procedures 1 Rinse all burets and flasks with distilled water. Discard wash in sink 2 Rinse buret with 5 ml of Base Discard in waste container Fil buret with Base. Record starting volume. 4. Add 10 ml of distilled water to cleaned flask 5. Pipet exactly 100 ml of acid to flask with water. Add 2 drops of Indicator 7 Add NaOH base slowly. Swirl Add base until 1 drop turns the solution very faint pink Record the volume of the base. 9. De another trial. Use the ending value of the base as the new trial's starting value 10 Add water, add and indicator to flask as before Titrate to a faint pink. Record volumes 12 De a third trial 1.20 M HCI Concentration This is sample data from a previous lab Data Table: Trial 1 Trial 2 HCI ml 10.00 ml 10.00ml Start NaOH ml 0.00 ml 8.31 ml End NaOH ml 8.31 ml 15.95 ml used NaOH ml 8.31 ml 7.64 ml Calculations: 1. Calculate the average ml NaOH used in the 3 trials Trial 3 10.00ml 15.95 ml 23.78 7.83 ml 2. Using stoichiometry for neutralization, calculate the Molarity of your NaOH solution. NaCl + H2O (start with HCI) HCI + NaOH 10.00 ml ave ml from #1 1.20M ?M 3. Using the M of NaOH from #2 as [OH-], what is the [H3O+] (use Kw=[H3O+](OH-), solve for [H3O+]) 4. What is the pH using the [H3O+] in #3 (use pH=-log [H3O+]