Respuesta :
The question is incomplete, here is the complete question:
A chemistry student weighs out 0.0975 g of acrylic acid [tex](HCH_2CHCO_2)[/tex] 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
Explanation:
To calculate the molarity of solution, we use the equation:
[tex]\text{Molarity of the solution}=\frac{\text{Mass of solute}\times 1000}{\text{Molar mass of solute}\times \text{Volume of solution (in mL)}}[/tex]
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:
[tex]\text{Molarity of acrylic acid}=\frac{0.0975\times 1000}{72\times 250}\\\\\text{Molarity of acrylic acid}=5.42\times 10^{-3}M[/tex]
To calculate the volume of base, we use the equation given by neutralization reaction:
[tex]n_1M_1V_1=n_2M_2V_2[/tex]
where,
[tex]n_1,M_1\text{ and }V_1[/tex] are the n-factor, molarity and volume of acid which is [tex]HCH_2CHCO_2[/tex]
[tex]n_2,M_2\text{ and }V_2[/tex] are the n-factor, molarity and volume of base which is NaOH.
We are given:
[tex]n_1=1\\M_1=5.42\times 10^{-3}M\\V_1=250mL\\n_2=1\\M_2=0.0500M\\V_2=?mL[/tex]
Putting values in above equation, we get:
[tex]1\times 5.42\times 10^{-3}\times 250.0=1\times 0.0500\times V_2\\\\V_2=\frac{1\times 5.42\times 10^{-3}\times 250}{1\times 0.000}=27.1mL[/tex]
Hence, the volume of NaOH needed is 271 mL
