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

Answer 1

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)