The number of moles of `Ca(OH)_(2)` required to prepare 250 ml of solution with pH 14 (assuming complete ionization) is

Options

A. `0.25`
B. `1.0`
C. `0.125`
D. `10.0`

Explanation

`C_(H^(+))=10^(-pH)mol L^(-1)`
`=10^(-14)mol L^(-1)`
For any aqueous solution at 298K, we have
`pH+pOH=14`
`:. pOH=14-pH=14-14=0`
Since `C_(OH^(-))=10^(-pOH)mol L^(-1)`
`=10^(o)mol L^(-1)`
`=1 mol L^(-1)`
According to dissociation equation,
`Ca(OH)_(2)rarr Ca^(2+)(aq.)+2OH^(-)(aq.)`
Every 1 mol of `Ca(OH)_(2)` provides 2 mol of `OH^(-)` ions. Thus,
`C_(Ca(OH)_(2))=(1)/(2)C_(OH^(-))=(1)/(2)(1 mol L^(-1))`
`=0.5 mol L^(-1)`
Now,
Molarity (M) `= (n_(Ca(OH)_(2)))/(V_(mL))xx(1000mL)/(L)`
`0.5mol L^(-1)= (n_(Ca(OH)_(2)))/(250 mL)xx(1000 mL)/(L)`
`= 0.125 mol`