Question

A 0.881 g sample of a diprotic acid is dissolved in water and titrated with 0.160 M NaOH . What is the molar mass of the acid if 35.8 mL of the NaOH solution is required to neutralize the sample

Answer 1

Answer: The molar mass of the acid if 35.8 mL of the NaOH solution is required to neutralize the sample is 308 g/mol

Explanation:

To calculate the number of moles for given molarity, we use the equation:

$\text{Molarity of the solution}=\frac{\text{Moles of solute}\times 1000}{\text{Volume of solution (in L)}}$     .....(1)

Molarity of NaOH solution = 0.160 M

Volume of solution = 35.8 mL

Putting values in equation 1, we get:

$0.160M=\frac{\text{Moles of NaOH}\times 1000}{35.8ml}\\\\\text{Moles of NaOH}=\frac{0.160mol/L\times 35.8}{1000}=0.00573mol$

$H_2A+2NaOH\rightarrow Na_2A+2H_2O$

As 2 moles of NaOH reacts with = 1 mole of diprotic acid

Thus 0.00573 moles of NaOH reacts with =$\frac{1}{2}\times 0.00573=0.00286$ moles of diprotic acid

mass of acid =$moles\times {\text {Molar mass of acid}}$

0.881 g =$0.00286\times {\text {Molar mass of acid}}$

${\text {Molar mass of acid}}=308g/mol$

Thus the molar mass of the acid if 35.8 mL of the NaOH solution is required to neutralize the sample is 308 g/mol

Answer 2

Answer:

No of moles = mass / molar mass

Molar mass = 0.881/0.260

Molar mass = 5.51M

Explanation:

A diprotic acid is titrated with NaOH solution of known concentration. Molecular weight (or molar mass) is found in g/mole of the diprotic acid. Weighing the original sample of acid will tell you its mass in grams. Moles can be determined from the volume of NaOH titrant needed to reach the first equivalence point.