Question

# What $[O{{H}^{-}}]$ in the final solution prepared by mixing 20 mL of 0.05M HCl with 30mL of 0.10M $Ba{{(OH)}_{2}}$ ?A. 0.12MB. 0.10MC. 0.40MD. 0.005M

Hint: The reactant that fully reacts in the reaction is called reactant limiting or reagent limiting. The reactant that is not fully consumed in the reaction is called excess reactant. Write the balanced reaction equation before solving the question. A balanced equation is an equation for a chemical reaction in which the number of atoms in the reaction for each element and the total charge for both the reactants and the products is equal.

The reaction used in question is:

$Ba{{(OH)}_{2}}+HCl\to BaC{{l}_{2}}+2{{H}_{2}}O$

Number of moles of ${{H}^{+}}$ ions present in solution = concentration of HCl $\times$ Volume of HCl
$=0.05\times \dfrac{20}{1000}=0.001$ moles
There are two $O{{H}^{-}}$ ions in one molecule of $Ba{{(OH)}_{2}}$.
$O{{H}^{-}}$ ions released in the Barium Hydroxide solution = concentration of $Ba{{(OH)}_{2}}$ $\times$ Volume of $Ba{{(OH)}_{2}}$ $\times$ 2.
$[O{{H}^{-}}]$ = $0.1\times \dfrac{30}{1000}\times 2=0.006$ moles.
${{H}^{+}}$ ion is limiting reagent.
Therefore, 0.001 moles of ${{H}^{+}}$ and $O{{H}^{-}}$ combine to form water.
0.005 moles of ions are $O{{H}^{-}}$ remaining.
Molarity is defined as the number of moles of solute dissolved in one litre (or one cubic decimeter) of solution. It is denoted by M.
Molarity, $M=\dfrac{n}{V}$ where, n is the number of moles and v is the volume of the solution.
Total volume of the solution = (20 + 30) mL = 50 mL = 0.05L.
Hence, the molarity of $O{{H}^{-}}$ ion = $\dfrac{0.005moles}{0.05L}$ = 0.100M

So, the correct answer is “Option B”.

Note: Don’t get confused between molarity and molality. Molarity (M) is defined as the number of moles of solute dissolved in one litre of solution whereas Molality (m) is defined as number of moles of solute per kilogram (kg) of the solvent.
For avoiding calculation mistake convert all quantities into SI unit before calculation