Question

What is the $\left[ {OH} \right]$ in the final solution prepared by mixing $20\,ml$ of $0.050\,M\,HCl$ with$30\,ml\,0.10\,M\,Ba{\left( {OH} \right)_2}$?
A) $0.12\,M$
B) $0.10\,M$
C) $0.40\,M$
D) $0.0050\,M$

Answer 1

Hint: We know that the number of equivalents per mole of an ion equals the charge on the ion and this leads to the definition of milliequivalent. Equivalent per liter is a unit of concentration $mEq/L.$ mathematically, mili equivalent is represented as,
$mEq = \dfrac{{\left( {mass} \right)\left( {Valence} \right)}}{{Molecular\,weight}}$
Example: We have to find the milliequivalent of potassium in $750\,ml$ of solution contain $58.65\,mg/L$ of potassium ion and the valence of potassium is $1$.Using the above equation, the milliequivalent of potassium is calculated as,
$mEq = \dfrac{{\left( {58.65} \right)\left( 1 \right)}}{{39.1}} = 1.5\,mEq$
We can calculate milliequivalent by using the following formula,
$mEq = Concentration \times No.of.ions \times volume$

Complete step by step answer:From the given data we can write the balanced equation as,
$Ba{\left( {OH} \right)_2} + 2\,HCl\xrightarrow{{}}BaC{l_2} + 2{H_2}O$
Given,
The volume of $HCl$ is $20\,ml.$
The molarity of $HCl$ is $0.050\,M$.
The volume of $Ba{\left( {OH} \right)_2}$ is $30\,ml$.
The molarity of $Ba{\left( {OH} \right)_2}$ is $0.1\,M$.
First, calculate the milliequivalent of hydrochloric acid and barium hydroxide.
The number of milliequivalent of hydrochloric acid can be calculated as,
The number of milliequivalent of hydrochloric acid $ = 20 \times 0.05 \times 1 = 1$
The number of milliequivalent of Barium hydroxide can be calculated as,
The number of milliequivalent of Barium hydroxide$ = 2 \times 30 \times 0.1 = 6$
Now, calculate the concentration of $OH$ using the equation.
$\left[ {O{H^ - }} \right] = \dfrac{{Milliequivalents\,Of\,Ba{{\left( {OH} \right)}_2} - Milliequivalents\,Of\,HCl}}{{Total\,volume}}$
$\left[ {O{H^ - }} \right] = \dfrac{{6 - 1}}{{30 + 20}}$
$\left[ {O{H^ - }} \right] = \dfrac{5}{{50}}$
$\left[ {O{H^ - }} \right] = 0.1\,M$
$\therefore $The concentration of $OH$ is $0.1\,M$.


Note:
 We can also calculate the concentration of $O{H^ - }$ as,
We know that,
\[Molarity = \dfrac{{No.of.\,moles}}{{Volume\,\left( L \right)}}\]
  The number of moles of hydrogen ion present in the solution is calculated as,
The number of moles of hydrogen \[ = Concentration{\text{ }}of\;HCl \times Volume\]
The number of moles of hydrogen \[ = \dfrac{{0.05\,mole}}{{ml}} \times 0.020ml = 0.001\,moles\]
The number of moles of hydroxyl ion \[ = Concentration{\text{ }}of\;Ba{\left( {OH} \right)_2} \times Volume\]
There are two hydroxyl ions in barium hydroxide.
The number of moles of hydroxyl ion \[ = \dfrac{{0.1mole}}{{ml}} \times 0.030ml \times 2 = 0.006\,moles\]
The number of moles of hydrogen is lower hence it is a limiting agent.
$\left[ {O{H^ - }} \right] = \dfrac{{Moles\,O{H^ - } - Moles\,of\,{H^ + }}}{{Total\,volume}}$
$\left[ {O{H^ - }} \right] = \dfrac{{0.006 - 0.001}}{{0.05L}} = 0.1\,M$
The concentration of $OH$ is $0.1\,M$.