Question

A sample of oxalic acid (a diprotic acid of the formula H2C2O4) is dissolved in enough water to make 1.00 L of solution. A 100.0 mL sample of this solution is titrated with a solution of sodium hydroxide of concentration 0.750 M and requires 20.0 mL of sodium hydroxide to reach the end point. Calculate the mass of the original oxalic acid sample.

Answer 1

Answer: The mass of original oxalic acid sample is 6.75 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 $H_2C_2O_4$

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

We are given:

$n_1=2\\M_1=?M\\V_1=100.0mL\\n_2=1\\M_2=0.750M\\V_2=20.0mL$

Putting values in above equation, we get:

$2\times M_1\times 100.0=1\times 0.750\times 20.0\\\\M_1=\frac{1\times 0.750\times 20.0}{2\times 100.0}=0.075M$

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

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

Given mass of oxalic acid = ? g

Molar mass of oxalic acid = 90 g/mol

Molarity of solution = 0.075 M

Volume of solution = 1.00 L

Putting values in above equation, we get:

$0.075M=\frac{\text{Mass of oxalic acid}}{90g/mol\times 1L}\\\\\text{Mass of oxalic acid}=(0.075\times 90\times 1)=6.75g$

Hence, the mass of original oxalic acid sample is 6.75 grams