Question

What is the $[O{{H}^{-}}]$ in the final solution prepared by mixing 20.0 ml of 0.050 M $HCl$ with 30.0 ml of 0.10 M $Ba{{(OH)}_{2}}$?
A. 0.12 M
B. 0.10 M
C. 0.40 M
D. 0.050 M

Answer 1

Hint: Solution is made up of two components known by the name solute and solvent where solute is that substance which gets dissolved and solvent in which solute is dissolved and they both make solution.

Complete answer:
To find the value of $[O{{H}^{-}}]$, we first have to write the reaction between $Ba{{(OH)}_{2}}$ with $HCl$ which is as follows:
$Ba{{(OH)}_{2}}+2HCl\to BaC{{l}_{2}}+2{{H}_{2}}O$
Number of moles of ${{H}^{+}}$ ions present in the solution can be calculated by multiplying the concentration of $HCl$ which is 0.05 given with volume of $HCl$i.e 20 ml but we have to calculate it in liters that would be:
$0.05\times \dfrac{20}{1000}=0.001$moles
We can see that there are two $O{{H}^{-}}$ ions in one molecule of $Ba{{(OH)}_{2}}$.
Therefore $O{{H}^{-}}$ can also be calculated by the same formula i.e. multiplying the concentration of $Ba{{(OH)}_{2}}$ which is 0.10 given with volume of $Ba{{(OH)}_{2}}$i.e 30 ml into 2 as 2 molecules of $Ba{{(OH)}_{2}}$ so it can be calculated as:
$0.1\times \dfrac{30}{1000}\times 2=0.006$moles
We know that ${{H}^{+}}$ ion is limiting reagent.
Therefore 0.001 moles of ${{H}^{+}}$ and $O{{H}^{-}}$ combine to form water and now we left with 0.005 moles of $O{{H}^{-}}$ ions.
Total volume of the solution = 20 + 30 = 50 ml = 0.05 L
So the molarity of $O{{H}^{-}}$ion = $\dfrac{Number\ \text{of moles}}{volume}=\dfrac{0.005}{0.05}=0.100M$

Hence option B is the correct answer.

Note:
Molarity of any solution is represented by the symbol M and known by the term and molarity can be defined as the total number of moles of solute dissolves in one liter of the solution. Molarity of a solution generally depends upon the volume of the solution.