Question

Find the pH of $0.005M$ $Ba{(OH)_2}$ solution at $25^\circ C$. Also calculate the pH value of the solution when $100ml$ of the above solution is diluted to $1000ml$. (Assume complete ionization of barium hydroxide).
A.$8,9$
B.$2,3$
C.$10,12$
D.$12,11$

Answer 1

Hint: The pH of a solution can be calculated by using the following formulae,
$pH = - \log [{H^ + }]$
$pOH = - \log [O{H^ - }]$
And $pH + pOH = 14$
The number of moles of $O{H^ - }$ in the initial solution divided by $1000ml$ (volume of diluted solution) will give the concentration of $O{H^ - }$ in the diluted solution.

Complete step by step answer:
In $0.005M$ $Ba{(OH)_2}$ solution, the reaction for ionization of $Ba{(OH)_2}$ is,
$Ba{(OH)_2} \rightleftharpoons B{a^{2 + }} + 2O{H^ - }$
$0.005M$ $Ba{(OH)_2}$ means there are $0.005$ moles of $Ba{(OH)_2}$ in $1000ml$ of solution.
Since, one mole of $Ba{(OH)_2}$ gives two moles of $O{H^ - }$ ions, therefore, $0.005$ moles of $Ba{(OH)_2}$ will give
$ = 0.005 \times 2$ moles of $O{H^ - }$
$ \Rightarrow 0.01$ moles of $O{H^ - }$
So, concentration of $O{H^ - }$ ions in $1000ml$ or $1L$ of solution
$ = \dfrac{{0.01moles}}{{1L}}$
$ \Rightarrow 0.01M$
Therefore, $pOH = - \log [0.01]$
$ \Rightarrow pOH = - ( - 2)$
$ \Rightarrow pOH = 2$
As, $pH = 14 - pOH$
$ \Rightarrow pH = 14 - 2$
$ \Rightarrow pH = 12$
There are $0.01$ moles of $O{H^ - }$ ions in $1000ml$ of initial solution. As we have taken $100ml$ of initial solution, so number of moles of $O{H^ - }$ ions are
$ = \dfrac{{0.01}}{{1000}} \times 100$
$ \Rightarrow 0.001$ moles
As the volume of solution is made $1000ml$, i.e. $1L$, so concentration of $O{H^ - }$ ions in the diluted solution is
$ = \dfrac{{0.001moles}}{{1L}}$
$ \Rightarrow 0.001M$
The pOH of diluted solution $ = - \log [0.001]$
$ \Rightarrow - ( - 3)$
$ \Rightarrow 3$
So, $pH = 14 - 3$
$ \Rightarrow pH = 11$.

Thus, the answer is option D.

Note:
In this question, for calculating the pH of the solution, instead of calculating the pOH of the solution and then subtracting it from $14$ to calculate the pH (using the formula $pH = 14 - pOH$). We can directly calculate the concentration of ${H^ + }$ ions from $O{H^ - }$ ions and then use it to calculate the pH by using the formula $pH = - \log [{H^ + }]$.
The concentration of ${H^ + }$ and $O{H^ - }$ ions are related as, $[{H^ + }][O{H^ - }] = {10^{ - 14}}$.