# homework help 19871

Titration is the addition of one solution of a known concentration (the titrant) to a known volume of another solution of unknown concentration (the analyte) until one has just reacted completely with the other.

Titration is used in many types of food analysis as well as in chemistry labs.

Here’s an example of how to calculate the percent of acetic acid in a sample of vinegar.

EXAMPLE

A 50.0 mL sample of vinegar was diluted to 250 mL. Titration of 20.0 mL of this solution required 28.0 mL of 0.100 mol/L NaOH. What was the mass percent of acetic acid in the vinegar? Assume that the vinegar has the same as water.

Solution

Step 1. Calculate the moles of NaOH.

Moles of NaOH = 0.0280 L NaOH × ##( 0.100″mol NaOH”)/(1″L NaOH”)## = 0.002 80mol NaOH

Step 2. Write the balanced chemical reaction for the reaction

CH₃COOH + NaOH → CH₃COONa + H₂O

Step 3. Calculate the moles of CH₃COOH.

Moles of acetic acid = 0.002 80 mol NaOH × ##(1″mol CH₃COOH”)/(1″mol NaOH”)## = 0.002 80 mol CH₃COOH

Step 4. Calculate the of the diluted vinegar.

Molarity = ##”moles”/”litres” = (0.002 80″mol”)/(0.0200″L”)## = 0.140 mol/L

Step 5. Calculate the molarity of the original vinegar.

##c_1V_1= c_2V_2##

##c_1 = c_2 × V_2/V_1## = 0.140 mol/L × ##(250″mL”)/(50.0″mL”)## = 0.700 mol/L

Step 6. Calculate the mass of CH₃COOH in the vinegar.

Mass of CH₃COOH = 0.0500 L vinegar ×

##(0.700″mol CH₃COOH “)/(1″L vinegar”) × (60.05″g CH₃COOH”)/(1″mol CH₃COOH”)## = 2.10 g CH₃COOH

Step 7. Calculate the mass percent of acetic acid.

Mass percent = ##”mass of CH₃COOH”/”total mass” × 100 % = (2.10″g”)/(50.0″g”)## × 100 % = 4.20 %

The vinegar contains 4.20 % acetic acid.

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