Question

The molecular mass of organic acid was determined by the study of its barium salt. 4.290 g of salt was quantitatively converted to the free acid by the reaction with 21.64 mL of 0.477 M ${ H }_{ 2 }{ SO }_{ 4 }$. The barium salt was found to have two moles of water of hydration per ${ Ba }^{ 2+ }$ ion and acid is monobasic. What is the molecular weight of anhydrous acid?
(a) 134
(b) 142
(c) 100
(d) 110

Answer 1

The barium salt reacts with the sulphuric acid such that it will give barium sulphate along with the monobasic acid from which the barium salt was composed of. Barium sulphate has a very low solubility product in water.

Complete step by step answer:
In order to solve this question let us first understand the meaning of the 0.477 M solution. M stands for molarity of the solution which can be written as mol/Litre. It is used in order to express the concentration of a solution and is equal to the number of moles of the solute per litre of the solution. The mathematical expression is given below:
$Molarity=\cfrac { Number of moles of solute }{ Volume of solution in Litres }$
Now let us solve the question:
It is given that the weight of the barium salt is 4.290 g and the volume of sulphuric acid (with the concentration of 0.477 M) that is used in order to convert the salt into a monobasic acid is 21.64 mL.
Therefore, reaction between the barium salt and the sulphuric acid will result in the formation of barium sulphate along with the monobasic salt. The reaction is given below:
$(RCOO{ ) }_{ 2 }Ba.{ 2H }_{ 2 }O(aq)+{ H }_{ 2 }{ SO }_{ 4 }(aq)\rightarrow { BaSO }_{ 4 }(s)\downarrow +2RCOOH+2{ H }_{ 2 }O(l)$

Where $(RCOO{ ) }_{ 2 }Ba.{ 2H }_{ 2 }O$ is the barium salt containing two moles of water of crystallization. The salt barium sulphate is insoluble in water due to the very small enthalpy of solvation of the barium and sulphate ions due to their large size which is not able to overcome the high lattice enthalpy and therefore it precipitates out.
Now we need to find out the number of moles of sulphuric acid that are reacting with the barium salt.
In a 0.477 M sulphuric acid solution:
1000 mL of the sulphuric acid solution contains = 0.477 moles of the sulphuric acid
Therefore, 21.64 mL of the sulphuric acid solution will contain = $\cfrac { 0.477 mole\times 21.64 mL }{ 1000 mL } =0.01 mole$

Now, from the chemical reaction we can see that 1 mole of the sulphuric acid reacts with 1 mole of the barium salt. Therefore 0.01 mole of the sulphuric acid will react with 0.01 mole of the barium salt. Hence the molecular weight of the barium salt will be:
Molecular weight = $\cfrac { 4.290 g }{ 0.01 mol } =429 g/mol$

This is the molecular weight of the barium salt containing the water of crystallization. To find the molar mass of the monobasic acid we need to subtract the mass of two moles of water along with the mass of barium from 429 g and add a mass of 2 g to the resultant answer for the mass of two protons.
$429 g/mol-127 g/mol(mass of barium ion)-(2\times 18 g/mol)(mass of water)$ = 266 g/mol + 2 g/mol = 268 g/mol
Now we need to divide this mass by two in order to get the mass for 1 mole of the monobasic acid:
$\cfrac { 268 g/mol }{ 2 } =134 g/mol$
So, the correct answer is “Option A”.

Note: Apart from molarity, there are many other terms that are used in order to express the concentration for a solution. Such as molality which is defined as the number of moles of the solute per kilogram of the solvent.