Question

Find the $pH$ of $0.005MBa{\left(OH\right)}_2$ solution at $25℃$. 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

The correct option is D $12,11$

$Ba{\left(OH\right)}_2⟶{Ba}^{2+}+2{OH}^{-}$

Initial$-0.005M$ $0$ $0$

$0$ $0.005M$ $2\times 0.005M$

$\therefore \left[{OH}^{-}\right]=0.01M$

$\therefore pOH=-\log \left[{OH}^{-}\right]=-\log \left(0.01\right)=2$

But $pH+pOH=14$

$\therefore pH=12$

$\therefore {\left[H^{+}\right]}_1={10}^{-pH}={10}^{-12}M$

Now, $M_1{V}_1=M_2{V}_2$

${10}^{-12}\times 100=M_2\times 1000$

$\therefore M_2={10}^{-11}M$

$\therefore {\left[H^{+}\right]}_2={10}^{-11}M$

$pH$ of solution when diluted $=-\log \left[{10}^{-11}\right]$

$=11$