Question

Find the $pH$ of $0.005M$ $Ba{\left(OH\right)}_2$ solution at $25$. Also calculate the $pH$ value of the solution when $100$ml of the above solution is diluted to $1000$ml. (Assume complete ionization of barium hydroxide)

• A. $8,9$
• B. $2,3$
• C. $10,12$
• D. $12,11$

Answer 1

Answer: (D)

$12,11$

$Ba(OH{)}_2⟶B{a}^{+2}+2O{H}^{-}$

at $t=0$ $0.005$ - -

at $t=t$ $0.005-0.005$ $0.005$ $2\times 0.005$

$\left[O{H}^{-}\right]=0.01$

$\Rightarrow p\left[OH\right]=-\log \left[O{H}^{-}\right]$

$\Rightarrow pOH=-\log {10}^{-2}=2$

$\Rightarrow pH+pOH=14$

Thus $pH=14-2=12$

On dilution of the solution, the new concentration of $Ba(OH{)}_2$ will be

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

$\Rightarrow 0.005\times 100=M_2\times 1000$

$\Rightarrow M_2=5\times {10}^{-4}M⟶$ Concentration of $\left[Ba\right(OH{)}_2]$

$Ba(OH{)}_2⟶B{a}^{+2}+2O{H}^{-}$

at $t=0$ $5\times {10}^{-4}M$ - -

at $t=t$ $0$ $5\times {10}^{-4}M$ $2\times 5\times {10}^{-4}M$

${[}^{-}OH]={10}^{-3}M$

$pOH=3$

$pH=14-3=11$ .