Now we know that #pK_a+pK_b=14# for an aqueous solution under standard conditions, #K_b=2xx10^-5#, and thus #pK_b=4.70#, and #pK_a=9.30#
And we substitute the #pK_a# value back into the original equation, with the desired #pH#.
And thus,
#9=9.30+log_10(([NH_4^+])/([NH_3]))#
And thus #-0.30=log_10(([NH_4^+])/([NH_3]))#
#10^(-0.3)=([NH_4^+])/([NH_3])#
And we know that #[NH_3]=0.50*mol*L^-1#.
So #[NH_4^+]=10^(-0.3)xx0.50*mol*L^-1=0.25*mol*L^-1#
Now #"concentration"# #=# #("Moles of solute")/("Volume of solution")#, but the volume of solution was SPECIFIED to be #20xx10^-3L#.
So #"moles of solute"=20xx10^-3cancelLxx0.25*mol*cancel(L^-1)#
#=5.01xx10^-3*mol# of #NH_4Cl#; and this represents a mass of
#5.01xx10^-3*molxx53.49*g*mol^-1=0.268*g#.
So a final check,
#[NH_4^+]=((0.268*g)/(53.49*g*mol^-1))/(20xx10^-3*L)=0.251*mol*L^-1#.
#[NH_3]=0.50*mol*L^-1#.
Resubstituting into the buffer equation:
#pH=pK_a+log_10(((0.251*mol*L^-1))/((0.50*mol*L^-1)))#
#=9.30+(-0.3)#
#=9.0# AS REQUIRED....................