1- A group of Chemistry Lab Students had the follwing data for three Trials of T

Question

1- A group of Chemistry Lab Students had the follwing data for three Trials of Titration of 20.00ml of grocery grade vinegar (acetic acid) with 0.500M NaOH:

Trial One: 34.4ml,   Trial Two:   34.2 ml, Trial Three: 34.5 ml.   What was the average # ml of 0.500 M NaOH needed to completely neutralize the 20.00 ml of vinegar?

2- Write the balanced Equation for the reaction of HC2H3O2 (acetic acid) with NaOH:

3-From your answer to Question #1, calculate how many moles of NaOH the students used

4-From your answer to Question #3, calculate how many moles of HC2H3O2 the students used

5-From your answer to Question #4, and the original volume of the vinegar used, calculate the molarity of the HC2H3O2 solution

Explanation / Answer

1- A group of Chemistry Lab Students had the follwing data for three Trials of Titration of 20.00ml of grocery grade vinegar (acetic acid) with 0.500M NaOH:

Trial One: 34.4ml,   Trial Two:   34.2 ml, Trial Three: 34.5 ml.   What was the average # ml of 0.500 M NaOH needed to completely neutralize the 20.00 ml of vinegar?

Average volume of NaOH used=(trial1volume +trial2volume+ trial3volume)/3=(34.4+34.2+34.5)/3=34.367mL

2- Write the balanced Equation for the reaction of HC2H3O2 (acetic acid) with NaOH:

CH3COOH + NaOH ===>CH3COONa + NaOH

3-From your answer to Question #1, calculate how many moles of NaOH the students used

Moles of NaOH used=molarity*Volume=0.50moles/L*0.03437L=0.0172moles of NaOH

4-From your answer to Question #3, calculate how many moles of HC2H3O2 the students used

since from balanced equation 1mole of NaOH reacts completely with=1mole CH3COOH

moles of CH3COOH=moles of NaOH=0.0172moles

5-From your answer to Question #4, and the original volume of the vinegar used, calculate the molarity of the HC2H3O2 solution

Volume of CH3COOH=20mL=0.020L

Molarity of CH3COOH=moles/Volume=0.0172moles/0.020L=0.859M

Related Questions

Navigate

• Browse (All)
• Browse 1
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.