The pH an aqueous solution of `Ba(OH)_(2)` is `10.0`. If the `K_(sp)` of `Ba(OH)_(2)` is `1.0xx10^(-9)`, the concetration of `Ba^(2+)` ions in the solution is

Options

A. `1.0xx10^(-5)M`
B. `1.0xx10^(-1)M`
C. `1.0xx10^(-4)M`
D. `1.0xx10^(-2)M`

Answer

B

Explanation

Unless otherwise stated, the solvent is water and the temperature is `25^(@)C`.
`Ba(OH)_(2)(s)hArrBa^(2+)(aq.)+2OH^(-)(aq.)`
`pH=10.0`
Since `pH+pOH=14.0(at 298 K)`
`pOH= 14.0-10.0`
`= 4.0`
`:. C_(OH^(-))= 10^(-4)mol L^(-1)`
According to solubility equilibrium,
`K_(sp)=C_(Mg^(2+))C_(OH^(-))^(2)`
`1.0xx10^(-9)=C_(Mg^(2+))(10^(-4))^(2)`
`C_(Mg^(2+))=(1.0xx10^(-9))/(10^(-8))`
`= 1.0xx10^(-1)M`