Question

Exact molarity of 1M NaOH used is 1.083M, the volume of NaOH used is 10.0mL, the final volume of stock NaOH is 100.00mL, and the molarity of stock NaOH solution is 0.1083M.

KHP had a molar mass of 204.2 g/mole.

B. Titration of KHP 1. 2. 3. Mass of KHP Initial buret reading Final buret reading Trial 1 Trial 2 0.30T usly & 0.00ml D.DD_ml 15.3 L 142 ml t is 1/0 ml _mol mol mol Trial 3 0.308 0.01_ml 15.12 Trial 4 0.3038 6.0_ml. 14.99 mL 14.99 mL "V final - Vinitial ar * 15.2 5. Moles KHP used mol mol _mol 6. Moles NaOH used mol mo! IT 7. Molarity of NaOH M 8. Average Molarity Calculations: moles: musicale (1029) 1. How does your diluted molarity (Part A) compare to your standardized value (Part BY? Are the values close? Which is more precise and why?

acid base titration, part A

Answer 1

Completing table in Part B (ignoring trial 2) -

Moles of KHP = mass of KHP (g) / molar mass of KHP (g/mol)

Moles of NaOH = Moles of KHP (since stoichiometric ratio is 1:1)

Molarity of NAOH = Moles of NaOH / volume of NaOH (L)

example for trial 1 :

Moles of KHP = 0.307/204.2 = 1.503*10^-3 moles

Moles of NaOH = 1.503*10^-3 moles

Molarity of NaOH = (1.503*10^-3) / (15.31*10^-3) = 0.098M

TRIAL 1 0-3014 TRIAL 4 0.3059 TRIM 3 కంకి MaKMP 15. 22ml Velume NabHued 15.31ml 14 94 md 3 MolsKut ued 1-484 K10 1-503 x/0-3 1.508 x/D 3 Na DH used Molis 3 1481x10 1-508 X10 1-503 X/03 Modandy g Naou 0.0189 M O.098 M O.099 M 0.09 86M Ayy Modarity

The diluted molarity (part A) is 0.1083M whereas the standardised molarity is 0.0986M . The difference in the two values is 0.0097M.

The standardised value is more precise since it is repeated 3 times to make it statistically valid.