SECTION PARTNER GRADE eriment 22 REPORT SHEET Titration 1. Volume of vinegar sam

Question

SECTION PARTNER GRADE eriment 22 REPORT SHEET Titration 1. Volume of vinegar sample 2. Molarity of NaOH solution 3. Initial reading of NaOH in buret 4. Final reading of NaOH in buret 5. Volume of NaOH used in titration: Trial 1 Trial 2 3.2 mL 2149mL 18.20 mL 113 mL .340 mot 0.3412 mot 6. Molarity of acetic acid in vinegar: (2) × [(5n1)] 7. Average molarity of acetic acid 8. Percent (w/v) of acetic acid in vinegar: 5.21 % /100 m L = (7) moL/1000 mL x 60 g/mel. x 0.1 x 1000 mN100 mL %=(7) x 60 × 0.1 |& w/v POST-LAB QUESTIONS 1. Assume that your vinegar contained a small amount of citric acid (a triprotic acid). Using the same experimental data, would you expect the molarity of this sample to be the same as or different than a sample which contained only pure acetic acid? 2. Assume that the tip of your buret was not properly filled with NaOH solution. It contained an air bubble which was eliminated during the titration. Would the calculated molarity of the vinegar be smaller, larger, or the same as the true molarity? Explain. Experiment 22 223

Explanation / Answer

1. triprotic acid will has more protons and will consume NaOH to neutralize. the molarity of the sample will increase

M1V1=M2V2

M2= M1V1/V2

so if, V1 increases the whole term increases

2.although a drop would not make much of a difference but if you see on a very precise scale then, again the molarity will increase because the reading of your biuret will be larger than the actual amount of NaOH added.

3.you can ignore this volume because the water is not going to consume NaOH, so the vinegar is just diluted but wiill still take up the same amount of base!

4. 5.21g w/v means 5.21g in 100ml

so in 250ml= 13.025g

5.vol of NaOh needed= 41.8-23.1= 18.7ml

molarity og NaOH= 0.2M

vol of HCOOH= 10ml

molarity of HCOOH=> (18.7* 0.2)/10= 0.374M

%wlv= molarity* molar mass *0.1= 0.374 *46= 17.2%

6. formic acid n= 5/46=0.11

molarity in 5ml = 0.11/0.1= 1.1M

M2 molarity of NaOH= 0.1868

M1V2=M2V2

V2= M1V1/M2

= 1.1*5/0.1868

= 29.44ml

acetic acid 5%- 5g in 100 ml water

n= 5/60=0.083

M1= 0.083/0.1= 0.83M

V1=5ml

M2= 0.1868M

V2= M1V1/M2

= 0.83*5/0.1868

=22.21ml

so, in your experiment you needed 17,36 ml (avg) of NaOH.

in this you will need more. 22.21-17.36= 4.85ml more

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