We use the buffer equation.......which is derived here, https://socratic.org/questions/how-do-buffers-maintain-ph#270129.
But here [A^-]=[NH_3][A−]=[NH3], and [HA]=[NH_4Cl][HA]=[NH4Cl].
For such a buffer pH=pK_a+log_10{[[NH_3(aq)]]/[[NH_4Cl(aq)]]}pH=pKa+log10{[NH3(aq)][NH4Cl(aq)]}
And I also ASSUME, that the required pH=8.90pH=8.90
Now pK_a=9.24pKa=9.24 for ammonium ion.
And thus, substituting these values into the equation..........
log_10{[[NH_3(aq)]]/[[NH_4Cl(aq)]]}=-0.34log10{[NH3(aq)][NH4Cl(aq)]}=−0.34
i.e. ([NH_3(aq)])/([NH_4Cl(aq)])=10^(-0.34)[NH3(aq)][NH4Cl(aq)]=10−0.34,
i.e. "moles of ammonium chloride"xx0.457="moles of ammonia"moles of ammonium chloride×0.457=moles of ammonia, BECAUSE the volume was the same in each case.
"moles of ammonia"=0.250*Lxx0.200*mol*L^-1=5.00xx10^-2*molmoles of ammonia=0.250⋅L×0.200⋅mol⋅L−1=5.00×10−2⋅mol
"ammonium chloride"=(5.00xx10^-2*mol)/(0.457)=0.109*molammonium chloride=5.00×10−2⋅mol0.457=0.109⋅mol
And thus we have to add............
0.109*molxx53.49*g*mol^-1=5.85*g0.109⋅mol×53.49⋅g⋅mol−1=5.85⋅g
To the initial 250.0*mL250.0⋅mL volume of ammonia.
Just as a recheck, let us substitute these values back into the buffer equation:
pH=9.24+log_10{{5.00xx10^-2*molxx1/0.2500*L)/((5.85*g)/(53.49*g*mol^-1)xx1/0.2500*L)}pH=9.24+log10⎧⎪⎨⎪⎩5.00×10−2⋅mol×10.2500⋅L5.85⋅g53.49⋅g⋅mol−1×10.2500⋅L⎫⎪⎬⎪⎭
=8.90=8.90 as required..........
