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.

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

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