25 mL of a solution of barium hydroxide on titration with 0.1 molar solution of hydrochloric acid gave a titre value of 35 mL. The molarity of barium hydroxide solution was:
Answer & explanation
Correct answer: option 1
The correct answer is option 1. 0.07.
To find the molarity of the barium hydroxide (\(Ba(OH)_2\)) solution, we can use the concept of stoichiometry and the equation:
\[ \text{Molarity} = \frac{\text{Moles of solute}}{\text{Volume of solution in liters}} \]
Given:
Volume of \(Ba(OH)_2\) solution = 25 mL = 0.025 L
Volume of \(HCl\) solution used in titration = 35 mL = 0.035 L
Molarity of \(HCl\) solution = 0.1 M
From the balanced chemical equation for the reaction between \(Ba(OH)_2\) and \(HCl\):
\[ Ba(OH)_2 + 2HCl \rightarrow BaCl_2 + 2H_2O \]
We can see that 1 mole of \(Ba(OH)_2\) reacts with 2 moles of \(HCl\).
Using the concept of equivalence point in the titration:
\[ \text{Moles of } Ba(OH)_2 = \text{Moles of } HCl \]
Let's calculate the moles of \(HCl\) first:
\[ \text{Moles of } HCl = \text{Molarity} \times \text{Volume} = 0.1 \, \text{M} \times 0.035 \, \text{L} = 0.0035 \, \text{moles} \]
Since the stoichiometric ratio between \(Ba(OH)_2\) and \(HCl\) is 1:2, the moles of \(Ba(OH)_2\) will be:
\[ \text{Moles of } Ba(OH)_2 = \frac{0.0035 \, \text{moles}}{2} = 0.00175 \, \text{moles} \]
Now, let's calculate the molarity of the \(Ba(OH)_2\) solution:
\[ \text{Molarity} = \frac{\text{Moles of } Ba(OH)_2}{\text{Volume of solution in liters}} = \frac{0.00175 \, \text{moles}}{0.025 \, \text{L}} = 0.07 \, \text{M} \]
So, the molarity of the barium hydroxide solution is \(0.07 \, \text{M}\).Therefore, the correct answer is (1) \(0.07\)