Question

The solubility of$Ba{\left( {OH} \right)_2}.8{H_2}O$ in water $288K$is$5.6g$ per $100g$ of water . What is the molarity of the hydroxide ions in saturated solution of $Ba{\left( {OH} \right)_2}.8{H_2}O$at $288K$?
A.$0.869$
B.$1.32$
C.$0.355$
D.None of these

Answer 1

As we know what is the molarity now we have to find here only the molarity of the hydroxide ions and we need to properly calculate it along with the number of ions attached. We know that the molarity is the number of moles of solute to the volume of the solution in litres

Complete step-by-step answer: Molarity is defined as the number of moles of solute to the volume of the solution in litres
Then we will calculate the molecular mass of the $Ba{\left( {OH} \right)_2}$which turns out to be
$315.463g/mol$.
Further we will talk about the given quantities in the question then we are given that it is $5.6g$ per $100g$ of water.
Now further we will calculate the number of moles in kilograms of water and on substituting the values we found that it is :
$\frac{{56}}{{315.4639}} = 0.1775$
Now we discuss the dissociation of $Ba{\left( {OH} \right)_2}.8{H_2}O$ as we know that dissociation is the splitting up of the compound into smaller ions or particles. This dissociation can be shown by the below equation:
$Ba{\left( {OH} \right)_2}.8{H_2}O \to B{a^{2 + }} + 2O{H^ - } + 8{H_2}O$
From this equation we get that that one mole of $Ba{\left( {OH} \right)_2}$ produces two moles of hydroxide ions
So we can multiply the number of moles of hydroxide ions by two
$2 \times 0.1775 = 0.355$moles
So we get the number of moles of hydroxide ions to be $0.355$moles.

Hence,option third that is option C is correct.

Note: Important point to note here is that we are not talking about the molarity of wentire compound, basically we are only talking about the barium hydroxide ions . Therefore the number of ions should be noted very carelly.