Question

A chemistry student weighs out of acrylic acid into a volumetric flask and dilutes to the mark with distilled water. He plans to titrate the acid with 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

The question is incomplete, here is the complete question:

A chemistry student weighs out 0.0975 g of acrylic acid
`(HCH_2CHCO_2)` into a 250. mL volumetric flask and diluted to the mark with distilled water. He plans to titrate the acid with 0.0500 M NaOH solution. Calculate the volume of NaOH solution the student will need to add to reach the equilvalence point. Be sure your answer has the correct number of significant digits.

Answer: The volume of NaOH needed is 27.1 mL

Step-by-step explanation:

To calculate the molarity of solution, we use the equation:

`\text{Molarity of the solution}=\frac{\text{Mass of solute}* 1000}{\text{Molar mass of solute}* \text{Volume of solution (in mL)}}`

Given mass of acrylic acid = 0.0975 g

Molar mass of acrylic acid = 72 g/mol

Volume of solution = 250 mL

Putting values in above equation, we get:

`\text{Molarity of acrylic acid}=(0.0975* 1000)/(72* 250)\\\\\text{Molarity of acrylic acid}=5.42* 10^(-3)M`

To calculate the volume of base, we use the equation given by neutralization reaction:

`n_1M_1V_1=n_2M_2V_2`

where,

`n_1,M_1\text{ and }V_1` are the n-factor, molarity and volume of acid which is
`HCH_2CHCO_2`

`n_2,M_2\text{ and }V_2` are the n-factor, molarity and volume of base which is NaOH.

We are given:

`n_1=1\\M_1=5.42* 10^(-3)M\\V_1=250mL\\n_2=1\\M_2=0.0500M\\V_2=?mL`

Putting values in above equation, we get:

`1* 5.42* 10^(-3)* 250.0=1* 0.0500* V_2\\\\V_2=(1* 5.42* 10^(-3)* 250)/(1* 0.000)=27.1mL`

Hence, the volume of NaOH needed is 271 mL