Question

A chemistry student weighs out 0.120 g of acetic acid into a 250. mL volumetric flask and dilutes to the mark with distilled water. He plans to titrate the acid with 0.0700 M NaOH solution. Calculate the volume of solution the student will need to add to reach the equivalence point. Be sure your answer has the correct number of significant digits.

Answer 1

Answer:

28.6 mL

Explanation:

Step 1: Write the balanced neutralization reaction between acetic acid and sodium hydroxide

CH₃COOH(aq) + NaOH(aq) = CH₃COONa(aq) + H₂O(l)

Step 2: Calculate the moles of acetic acid

The molar mass of acetic acid is 60.05 g/mol. The moles corresponding to 0.120 g are:

$0.120g \times \frac{1 mol}{60.05g} =2.00 \times 10^{-3} mol$

Step 3: Calculate the moles of sodium hydroxide

The molar ratio of CH₃COOH to NaOH is 1:1. The reacting moles of NaOH are 2.00 × 10⁻³ mol.

Step 4: Calculate the volume of the 0.0700 M NaOH solution

$2.00 \times 10^{-3} mol \times \frac{1L}{0.0700mol} =0.0286 L = 28.6 mL$

Answer 2

Answer:

We have to add 286 mL of NaOH

Explanation:

Step 1: Data given

Mass of acetic acid (CH3COOH)= 0.120 grams  

Volume of acetic acid = 250 mL = 0.250 L

Molarity of NaOH = 0.0700 M

Step 2: The balanced equation

CH3COOH + NaOH → CH3COONa + H2O

Step 3: Calculate moles acetic acid

Moles acetic acid = mass / molar mass

Moles acetic acid = 0.120 grams / 60.05 g/mol

Moles acetic acid = 0.00200 moles

Step 4: Calculate molarity acetic acid

Molarity acetic acid = moles / volume

Molarity acetic acid = 0.00200 moles /0.250 L

Molarity acetic acid = 0.008 M

Step 5: Calculate the volume of solution the student will need to add to reach the equivalence point

C1*V1 = C2*V2

⇒with C1 = the molarity of acetic acid = 0.008 M

⇒with V1 = the volume of acetic acid = 0.250 L

⇒with C2 = the molarity of NaOH = 0.0700 M

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

0.008 M * 0.250 L = 0.0700 M * V2

V2 = (0.008M * 0.250 L) / 0.0700 M

V2 = 0.286 L = 286 mL

We have to add 286 mL of NaOH