Question

The $pH$ of an aqueous solution of $Ba ( OH )_2$ is $10$ . If the $K_{ sp }$ of $Ba ( OH )_2$ is $1 \times 10^{-9}$, then the concentration of $Ba ^{2+}$ ions in the solution in $mol\, L ^{-1}$ is

• A. $1 \times 10^{-2}$
• B. $1 \times 10^{-4}$
• C. $1 \times 10^{-1}$
• D. $1 \times 10^{-5}$

Answer 1

Answer: (C)

$ pH =10, pOH =4, \stackrel{\ominus}{ OH }]=10^{-4} M .$
$Ba ( OH )_2 \longrightarrow Ba ^{2+}+2 \overset{\ominus}{ O}H $
${[\overset{\ominus}{ O}H ]=10^{-4}}$
(Since total concentration of $\overset{\ominus}{ O } H =10^{-4}$ )
$\therefore K_{ sp }=\left[ Ba ^{2+}\right][\overset{\ominus}{ O}H ]^2$
$10^{-9}=x \times\left(10^{-4}\right)^2 $
$\therefore x=10^{-1}=\left[ Ba ^{2+}\right]$