Question

A sample of 0.2140 g of an unknown monoprotic acid was dissolved in 25.0 mL of water and titrated with 0.0950MNaOH. The acid required 27.4 mL of base to reach the equivalence point. What is the molar mass of the acid?

Answer 1

To find the molar mass of the acid, we use the given data to calculate the moles of NaOH used and assume the reaction is 1:1 between the acid and NaOH. Using this information, we can then calculate the molar mass of the acid.

Step-by-step explanation:

To find the molar mass of the acid, we first need to calculate the number of moles of acid that reacted with the NaOH. The balanced chemical equation for the reaction is:

acid + NaOH → salt + water

From the given information, we know that 27.4 mL of 0.0950 M NaOH was required to react with 0.2140 g of the acid. Using this data, we can find the number of moles of NaOH used:

moles of NaOH = volume of NaOH (L) × concentration of NaOH (M)

moles of NaOH = 0.0274 L × 0.0950 M = 0.002606 mol NaOH

Since the reaction is 1:1 between the acid and NaOH, the number of moles of acid is also 0.002606 mol. Now we can calculate the molar mass of the acid:

molar mass = mass of acid (g) / moles of acid

molar mass = 0.2140 g / 0.002606 mol = 82.11 g/mol

Answer 2

in a monoprotic acid , its capable of dissociating 1 H⁺ ion . The base used is NaOH which is capable of giving out 1 OH⁻ ion.

Therefore both base and acid react in a 1:1 molar ratio, that means for each 1 mol of monoprotic acid there is , it requires 1 mol of NaOH to be neutralised.

So at equivalence point the number of acid moles present should react with an equal number of NaOH moles to be neutralised

number of NaOH moles added - concentration x volume

number of NaOH moles - 0.0950 mol/dm³ x 27.4 x 10⁻³ dm³ = 0.00260 mol

since molar ratio of NaOH to Acid is 1:1

number of acid moles present is also - 0.00260 mol

mass of acid added - 0.2140 g

we can find molar mass using following equation

molar mass = mass / number of moles

molar mass = 0.2140 g / 0.00260 mol = 82.3 g/mol

therefore molar mass of monoprotic acid is 82.3 g/mol