Question

A chemistry student weighs out 0.0617g of acetic acid (HCH3CO2) into a 250.mL volumetric flask and dilutes to the mark with distilled water. He plans to titrate the acid with 0.1300 M NaOH solution.Calculate the volume of solution the student will need to add to reach the equivalence point. Round your answer to 3 significant digits.

Answer 1

We have to add 7.92 mL of NaOH

Step-by-step explanation:

Step 1: Data given

Mass of acetic acid = 0.0617 grams

Volume of the flask = 250.0 mL = 0.250 L

Molarity of NaOH = 0.1300 M

Step 2: The balanced equation

CH3COOH + NaOH → CH3COONa + H2O

Step 3: Calculate moles acetic acid

Moles acetic acid = mass acetic acid / molar mass acetic acid

Moles acetic acid = 0.0617 grams / 60.05 g/mol

Moles acetic acid = 0.00103 moles

Step 4: Calculate concentration CH3COOH

Concentration CH3COOH = moles CH3COOH / volume

Concentration CH3COOH = 0.00103 moles / 0.250 L

Concentration CH3COOH = 0.00412 M

Step 5: Calculate the volume of NaOH needed

C1*V1 = C2*V2

⇒with C1 = the concentration of CH3COOH = 0.00412 M

⇒with V1 = the volume of CH3COOH = 0.250 L

⇒with C2 = the concentration of NaOH = 0.130 M

⇒with V2 = the volume of NaOH needed = TO BE DETERMINED

0.00412 M * 0.250 L = 0.130 M * V2

V2 = (0.00412 M * 0.250 L ) / 0.130 M

V2 = 0.00792 L = 7.92 mL

We have to add 7.92 mL of NaOH

Answer 2

7.92 mL

Step-by-step explanation:

Step 1: Write the balanced neutralization equation

HCH₃CO₂ + NaOH → NaCH₃CO₂ + H₂O

Step 2: Calculate the moles of acetic acid

The molar mass of acetic acid is 60.05 g/mol. The moles corresponding to 0.0617 g are:

`0.0617g * (1mol)/(60.05g) =1.03 * 10^(-3) mol`

Step 3: Calculate the moles of NaOH

The molar ratio of HCH₃CO₂ to NaOH is 1:1. The reacting moles of NaOH are 1.03 × 10⁻³ mol.

Step 4: Calculate the volume of NaOH

`1.03 * 10^(-3) mol * (1L)/(0.1300mol) =7.92 * 10^(-3) L = 7.92 mL`