Question

A solution of malonic acid, h2c3h2o4 , was standardized by titration with 0.1000 m naoh solution. if 21.55 ml of the naoh solution were required to neutralize completely 12.55 ml of the malonic acid solution, what is the molarity of the malonic acid solution?

Answer 1

The balanced chemical reaction between NaOH and malonic acid is as follows;
H₂C₃H₂O₄ + 2NaOH --> Na₂C₃H₂O₄ + 2H₂O
Stoichiometry of H₂C₃H₂O₄ to NaOH is 1:2
The number of moles of NaOH needed for complete neutralisation 
     - 0.100 mol/L /1000mL/L x 21.55 mL
Therefore number of NaOH moles that have reacted = 0.002155 mol 
For complete neutralisation , number of malonic acid moles = (number of NaOH moles )/ 2
malonic acid moles reacted = 0.002155/2 mol
Number of malonic acid moles in 12.55 mL = 0.002155/2 mol
Therefore in 1000 mL volume = (0.002155/2 mol)/12.55 mL x 1000 mL
Molarity of malonic acid = 0.086 mol/L

Answer 2

Answer:

0.0862M is  the molarity of the malonic acid solution.

Explanation:

$H_2C_3H_2O_4+2NaOH\rightarrow 2H_2O+Na_2C_3H_2O_4$

To calculate the concentration of acid, we use the equation given y 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 malonic acid.

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

We are given:

$n_1=2\\M_1=?M\\V_1=12.55mL\\n_2=1\\M_2=0.1000M\\V_2=21.55mL$

Putting values in above equation, we get:

$2\times M_1\times 12.5 mL=1\times 0.1000 M\times 21.55\\\\M_1=0.0862M$

0.0862M is  the molarity of the malonic acid solution.