Question

A 5.00g quantity of a diprotic acid was dissolved in water and made up exactly 250 mL. Calculate the molar mass if the acid is 25.0 mL of this solution required 11.1 mL of 1.00 KOH for neutralization. Assume both protons of the acid were titrated.

Answer 1

Answer:

The molar mass of the diprotic acid is 90.10 g/mol.

Explanation:

The acid is 25.0 mL of this solution required 11.1 mL of 1.00 KOH for neutralization.

$H_2A+2KOH\rightarrow K_2A+2H_2O$

To calculate the concentration of acid, we use the equation given by neutralization reaction:

$n_1M_1V_1=n_2M_2V_2$  ( neutralization )

where,

$n_1,M_1\text{ and }V_1$ are the n-factor, molarity and volume of diprotic acid

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

We are given:

$n_1=2\\M_1=?\\V_1=25 mL\\n_2=1\\M_2=1.00 M\\V_2=11.1 mL$

Putting values in above equation, we get:

$2\times M_1\times 25=1\times 1.00\times 11.1}$

$M_1=\frac{1\times 1.00\times 11.1}{2\times 25}=0.222 M$

Molarity of acid solution = 0.222 M =0.222 mol/L

Volume of original solution = 250 mL = 0.250 L ( 1 mL = 0.001 L)

Moles of diprotic acid in 0.250 L solution :

$=0.222 mol/L\times 0.250 L=0.0555 mol$

Mass of diprotic acid = m = 5.00 g

$Moles(n)=\frac{mass(m)}{\text{Molar mass(M)}}$

$M=\frac{m}{n}=\frac{5.00 g}{0.0555 mol}=90.10 g/mol$

The molar mass of the diprotic acid is 90.10 g/mol.