10% ammonia by mass means 10 g `NH_(3)` are present in 100 g of the solution.
`:.` Molarity of the solution `=(10)/(17) xx (1)/(100//0.99)xx1000 = 5.82 M`
`{:(,NH_(3) ,+,H_(2)O,rarr,NH_(4)OH,hArr,NH_(4)^(+),+,OH^(-),),("Initial conc.",,,,,C "mol" L^(-1),,,,,),("After dissociation",,,,,C-Calpha,,C alpha,,C alpha,),(,,,,,=C(1-alpha),,,,,):}`
`:. [OH^(-)] = C alpha = C sqrt((K_(b))/(C)) = sqrt(K_(b)C)= sqrt((K_(w))/(K_(a))xxC)=sqrt((10^(-14))/(5.0xx10^(-10))xx5.82)=1.079xx10^(-2)M (alpha = sqrt(K_(b)//C) and K_(a)xxK_(b)=K_(w))`
`:. [H^(+)]=(K_(w))/([OH^(-)]) = (10^(-14))/(1.079xx10^(-2))=0.9268xx10^(-12) M = 9.268xx10^(-13)M`