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Step-by-Step Solution
Step 1: Write the balanced chemical equation
The neutralization reaction between barium hydroxide and hydrochloric acid is:
$ \text{Ba(OH)}_{2} + 2\,\text{HCl} \to \text{BaCl}_{2} + 2\,\text{H}_{2}\text{O} $
Step 2: Identify known quantities
Volume of Ba(OH)2 solution $V_1 = 25\,\text{mL}$
Molarity of HCl $M_2 = 0.1\,\text{M}$
Volume of HCl $V_2 = 35\,\text{mL}$
Molarity of Ba(OH)2 = $M_1$ (to be determined)
Step 3: Relate moles of reactants using stoichiometry
From the balanced equation, $1$ mole of Ba(OH)2 reacts with $2$ moles of HCl. Hence, the relationship is:
$ \frac{M_1 \times V_1}{1000} \times 1 = \frac{M_2 \times V_2}{1000} \times \frac{1}{2}
$
The factor of 1000 converts mL to L, and the "1/2" on the right side accounts for the stoichiometric factor: 2 moles of HCl for every 1 mole of Ba(OH)2.
Step 4: Substitute values and solve for $M_1$
$\displaystyle \frac{M_1 \times 25}{1000} = \frac{0.1 \times 35}{1000} \times \frac{1}{2}
$
$\displaystyle M_1 \times 25 = 0.1 \times 35 \times \frac{25}{25}
$
Simplifying, we get:
$\displaystyle M_1 \times 25 = \frac{3.5}{2} = 1.75
$
$\displaystyle M_1 = \frac{1.75}{25} = 0.07
$
Step 5: State the final answer
The molarity of the Ba(OH)2 solution is $0.07\,\text{M}$.
