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.

Answer 2

The molar mass of the diprotic acid is 90.09 g/mol

From the question,

We are to determine the molar mass of the diprotic acid

First, we will write the balanced chemical equation for the reaction

Let the diprotic acid be H₂X

The balanced chemical equation for the reaction is

H₂X + 2KOH → K₂X + 2H₂O

This means,

1 mole of the diprotic acid is required to neutralize 2 moles of KOH

Now, we will determine the number of moles of KOH present

From the given information

Volume of KOH = 11.1 mL = 0.0111 L

Concentration of KOH = 1.00 M

Using the formula

Number of moles = Concentration × Volume

∴ Number of moles of KOH present = 1.00 × 0.0111

Number of moles of KOH present = 0.0111 mole

Since

1 mole of the diprotic acid is required to neutralize 2 moles of KOH

Then,

$\frac{0.0111}{2}$ mole of the diprotic acid would be required to neutralize the 0.0111 moles of KOH

$\frac{0.0111}{2}$ = 0.00555

∴ 0.00555 mole of the diprotic acid is required

Now, for the molar mass of the acid

Using the formula

$Molar\ mass = \frac{Mass}{Number\ of\ moles }$

From the given information

Mass of acid in the original solution = 5.00 g

If there are 5.00g of the acid in 250 mL

Then

There will be 0.500 g of the acid in 25.0 mL

∴ Mass of acid that reacted = 0.500 g

∴ Molar mass of the diprotic acid = $\frac{0.500}{0.00555}$

Molar mass of the diprotic acid = 90.09 g/mol

Hence, the molar mass of the diprotic acid is 90.09 g/mol

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