Question

You have a 2.0 mL sample of acetic acid (molar mass 60.05 g/mol) of unknown concentration. You titrate it to its endpoint with 20.0 mL NaOH (0.1 M). What mass of acetic acid was dissolved in the 2.0 mL of solution used?

Answer 1

Answer: The mass of acetic acid used is 0.12 grams

Explanation:

To calculate the concentration of acid, 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 $CH_3COOH$

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

We are given:

$n_1=1\\M_1=?M\\V_1=2.0mL\\n_2=1\\M_2=0.1M\\V_2=20.0mL$

Putting values in above equation, we get:

$1\times M_1\times 2.0=1\times 0.1\times 20.0\\\\M_1=\frac{1\times 0.1\times 20.0}{1\times 2.0}=1M$

To calculate the mass of solute, we use the equation used to calculate the molarity of solution:

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

Given mass of acetic acid = ? g

Molar mass of acetic acid = 60.05 g/mol

Molarity of solution = 1 M

Volume of the solution = 2.0 mL

Putting values in above equation, we get:

$1mol/L=\frac{\text{Mass of acetic acid}\times 1000}{60.05g/mol\times 2.0}\\\\\text{Mass of acetic acid}=\frac{1\times 60.05\times 2}{1000}=0.12g$

Hence, the mass of acetic acid used is 0.12 grams