Name Desk Date Laboratory Instructor REPORT SHEET EXPERIMENT Titration of Acids
ID: 551188 • Letter: N
Question
Name Desk Date Laboratory Instructor REPORT SHEET EXPERIMENT Titration of Acids and Bases A. Standardization of Sodium Hydroxide (NaOH) Solution Trial 1 Trial 2 Trial 3 Mass of bottle +KHP Mass of bottle Mass of KHP used Final buret reading Initial buret reading mL of NaOH used Molarity ofNaOH Average molaty D1DE Sdard deviation 125 10 Show your calculations for molaritý and standard deviation. 00 0.10 0.106M· 43. : 01010 0.0/9 B. Analysis of an Unknown Acid Trial I Trial 2 Trial 3 Mass of bottle + unknown Mass of bottle Mass of unknown used Final buret reading Initial buret reading mL of NaOH used Mass of KHP in unknown . 14 Percent of KHP in unknown % KWPExplanation / Answer
trial 1 :
moles of KHP used in the titration = 0.4190 gm/204.22 gmmol^-1 = 2.05*10^-3 moles
moles of NaOH = moles of KHP = 2.05*10^-3 moles
volume of NaOH used = 18 mL
molarity of NaOH = 2.05*10^-3 moles*1000mL/18mL = 0.113 M
trial 2 :
moles of KHP used in the titration = 0.426 gm/204.22 gmmol^-1 = 2.09*10^-3 moles
moles of NaOH = moles of KHP = 2.09*10^-3 moles
volume of NaOH used = 18 mL
molarity of NaOH = 2.09*10^-3 moles*1000mL/18.3mL = 0.114 M
trial 3 :
moles of KHP used in the titration = 0.439 gm/204.22 gmmol^-1 = 2.15*10^-3 moles
moles of NaOH = moles of KHP = 2.15*10^-3 moles
volume of NaOH used = 19.3 mL
molarity of NaOH = 2.15*10^-3 moles*1000mL/19.3mL = 0.111 M
Average = 0.113 M
Calculations for molarity and standard deviation:
M1=0.002mol/0.01950L=0.1026M
M2=0.002mol/0.01970L=0.1015M
M3=0.002mol/0.1980L=0.1010M
S=0.58(0.1026-0.1010)
=0.58*0.0016
=0.000928M
=9.28*10^-4M
B. Analysis of unknown acid :
trial1
moles of NaOH = 0.113 M *22.12mL/1000mL =2.5*10^-3 moles
moles of KHP = 2.5*10^-3 moles
mass of KHP = 2.5*10^-3 moles * 204.22 gm/mol = 0.510 gm
% = 0.510gm/0.99 *100 = 51.5 %
trial 2 :
moles of NaOH = 0.113 M *23.12mL/1000mL =2.61*10^-3 moles
moles of KHP = 2.61*10^-3 moles
mass of KHP = 2.61*10^-3 moles * 204.22 gm/mol = 0.533 gm
% = 0.533 gm/1gm *100 =53.3 5
trial 3:
moles of NaOH = 0.113 M *22.22mL/1000mL =2.51*10^-3 moles
moles of KHP = 2.51*10^-3 moles
mass of KHP = 2.51*10^-3 moles * 204.22 gm/mol = 0.512 gm
% = 51.2 %
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