Question

1. What is the molar mass of KHC,H04 to 4 significant digits (use two decimal places off of the periodic table)? Use the following data to complete the lab: Trial 1 Trial 2 Trial 3 Mass of KHC,H,04 () 1.19 1.16 1.22 0.20 1.52 Initial volume of NaOH 0.12 im Final volume of NaOH 27.36 26.50 29.28 1 a. Complete the following table for each trial. Trial 1 Trial 2 Trial 3 Mass of KHC3H04 () Moles of KHC HOA (mol) Initial volume of NaOH (mL) Final volume of NaOH (mL) Molar concentration of base 1 b. What is the average molarity of your standardized base? ar noss of a set of values obtained from identical murement

Answer 1

Titration of KHC8H4O4 with NaOH is as follows:--

KHC8H4O4 (aq) + NaOH(aq) ===> KNaC8H4O4 (aq) + H2O(l)

So according to the equation for complete titration

Moles of KHC8H4O4 = Moles Of NaOH

Moles of NaOH = Mass of KHC8H4O4/Molar mass of KHC8H4O4

Molar mass of KHC8H4O4 = 204.22 grams

Moles of NaOH = (Final volume - Intial Volume)/1000*Molarity

Trial 1

Moles of NaOH = Mass of KHC8H4O4/Molar mass of KHC8H4O4

Molar mass of KHC8H4O4 = 204.22 grams

Moles of KHC8H4O4 = 1.19/204.22

Moles of NaOH = (Final volume - Intial Volume)/1000*Molarity'

Moles of NaOH = (27.36-0.12)/1000*molarity

Moles of KHC8H4O4 = Moles Of NaOH

1.19/204.22 = (27.36-0.12)/1000*molarity

Molarity =  1.19/204.22/((27.36-0.12)/1000) M

= 0.214 M

Trial 2

Moles of NaOH = Mass of KHC8H4O4/Molar mass of KHC8H4O4

Molar mass of KHC8H4O4 = 204.22 grams

Moles of KHC8H4O4 = 1.16/204.22

Moles of NaOH = (Final volume - Intial Volume)/1000*Molarity'

Moles of NaOH = (26.50-0.20)/1000*molarity

Moles of KHC8H4O4 = Moles Of NaOH

1.16/204.22 = (26.50-0.20)/1000*molarity

Molarity =  1.16/204.22/((26.50-0.20)/1000) M

= 0.216 M

Trial 3

Moles of NaOH = Mass of KHC8H4O4/Molar mass of KHC8H4O4

Molar mass of KHC8H4O4 = 204.22 grams

Moles of KHC8H4O4 = 1.22/204.22

Moles of NaOH = (Final volume - Intial Volume)/1000*Molarity'

Moles of NaOH = (29.28-1.52)/1000*molarity

Moles of KHC8H4O4 = Moles Of NaOH

1.22/204.22 = (29.28-1.52)/1000*molarity

Molarity =  1.22/204.22/((29.28-1.52)/1000) M

= 0.215 M

1b.

Average molarity = 0.214+0.216+0.215/3

= 0.215 M