Molarity of NaOH solution: .25 M 1. Trial Titration Final buret volume (mL): 34.
ID: 955027 • Letter: M
Question
Molarity of NaOH solution: .25 M
1. Trial Titration
Final buret volume (mL): 34.4 mL
Initial Buret Volume (mL): 0mL
Volume of NaOH solution (mL): 34.4 mL
2. Exact titration
Sample no. 1 2 3 4
Final buret volume (mL)36.1, 34.9, 34.2, 46.2
intial buret buret volume (mL) 0, 0, 0, 12
Volume of NaOH solution (mL) 36.1, 34.9, 34.2, 34.2
concentration of HC2H3O2 (M) ________ ________ _________ _________
Mean Concentration (M) ________ ________ _________ _________
Show Calculations:
Questions:
1.The manufacture of the vnegar used in this experiment claims that the vinegar contains 5% acedic acid by weight. use your results and a density of 1.0 g/ml to determin of this claim is true or false
Explanation / Answer
1 M NaOH= 40 GMS IN 1000 ML ,or 4 gms in 100 ,or 0.4 in 10 ml
THEREFORE 0.25 M NaOH=0.25*40=10/1=10 gms NaOH Required FOR 1000 ML
1.Trail Titration= INITIAL READING=0 ML,final reading NaOH=34.4 ML
34.4 ML REQUIRED TO TITRATE ACETIC ACID ,
Now we can calculate the actual moles of NaOH FROM 34.4 ML .25 M CONC.
1000 ML =10 GMS THEN,
34.4 ML= ? BY CROSS MULTIPLYING,
34.4*10=344/1000=0.3440 GMS NaOH,
NOW WE CAN CALCULATE THE MOLES OF NaOH=0.3440/40=0.0086 moles,
for titration of 1 mole of acetic acid it is required 1 mole of NaOH ,
CH3COOH+ NaOH -----------> CH3COO-Na+ +H2O ,
Then for 0.0086 moles of NaOH IT IS REQUIRED 0.0086 MOLES OF acetic acid.
from that we can calculate the moles of acetic acid ,
that is conc. 0.0086*60=0.516 GMS acetic acid ,
2. 1=36.1 ml NaOH,
1000 ML =10 GMS THEN,
36.1 ML= ? BY CROSS MULTIPLYING,
36.1*10=344/1000=0.3610 GMS NaOH
MOLES OF NaOH=0.3610/40=0.0090 moles,
0.0090*60=.05415 GMS acetic acid,
2=34.9 ml NaOH,
1000 ML =10 GMS THEN,
34.9 ML= ? BY CROSS MULTIPLYING,
34.9*10=349/1000=0.3490 GMS NaOH
MOLES OF NaOH=0.3490/40=0.0087 moles,
0.0087*60=.05235 GMS acetic acid,
3=34.2 ml NaOH,
1000 ML =10 GMS THEN,
34.2 ML= ? BY CROSS MULTIPLYING,
34.2*10=342/1000=0.3420 GMS NaOH
MOLES OF NaOH=0.3420/40=0.00855 moles,
0.00855*60=0.513 GMS acetic acid,
4=FINAL READING - INITIAL READING=46.2-12=34.2 ML
1000 ML =10 GMS THEN,
34.2 ML= ? BY CROSS MULTIPLYING,
34.2*10=342/1000=0.3420 GMS NaOH
MOLES OF NaOH=0.3420/40=0.00855 moles,
0.00855*60=0.513 GMS acetic acid,
CONCENTRATION OF ACETIC ACIDS= (M)
1=0.0090,2=0.0087,3=0.00855,4=0.00855,
Mean concentration of acetic acid= 0.0090+0.0087+0.00855+0.00855=0.0348/4=0.0087 moles,
wt of acetic acids=0.5415+0.5235+0.513+0.513=2.091/4=0.522 grams (density of acetic acid =1.0 gm/ml)
this means that 0.522 gms in 10 ml, now calculate the % of acetic acid,
0.522/10=0.0522*100=5.22 % acetic acid , wt/v
% acetic acid = (wt/wt)=10 ml+0.522=10.522 total wt,
0.522/10.522=0.0496*100=4.96 %
claim by manufacturer is correct (5% vinegar)
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