An aqeous solution contains `10%` ammonia by mass and has density of `0.99 gm cm^(-3)`. The pH of this solution is
[Ka of `NH_(3) = 5 xx 10^(-10)M`]

Options

A. `11.033`
B. `12.033`
C. `9.033`
D. `7.2033`

Answer

B

Explanation

Given,
`("wt. of" NH_(3))/("wt. of solution") = (10)/(100)`
`100g` solution contains `10g NH_(3)`
`M_(NH_(3)) = ((10 xx 1000))/([17 xx (100//0.99)]) = 5.82`
`(':. V = ("mass")/("density"))`
Now,
`NH_(3) + H_(2)O rarr NH_(4)OH hArr NH_(4)^(+) + OH^(-)`
`"Before dissociation" ,1,0,0`
`"After dissociation",(1-Calpha),Calpha,Calpha`
`[OH^(-)] = C.alpha = Csqrt(((K_(b))/(C))) = sqrt((K_(b).C))`
`C = 5.82 M` and `K_(b) = K_(w)//K_(a)`
`= 10^(-14)//(5 xx 10^(-10)) = 2 xx 10^(-5)]`
`[OH^(-)] = sqrt([2 xx 10^(-5) xx 5.82]) = 1.07 xx 10^(-2) M`
`[H^(+)] = 10^(-14)//1.07 xx 10^(-2)`
`= 0.9268 xx 10^(-12) M`
`pH = - "log"[H^(+)] = - "log" 0.9268 xx 10^(-12) = 12.0330`.