Given #250*mL# of #0.200*mol*L^-1# aqueous ammonia, how much ammonium chloride would be added to achieve a #pH-=8.90#?

1 Answer
Apr 25, 2017

Answer:

Doubtless you mean #"ammonium chloride"#, #NH_4Cl#. I calculate a mass of #5.85*g#..................

Explanation:

We use the buffer equation.......which is derived here, https://socratic.org/questions/how-do-buffers-maintain-ph#270129.

But here #[A^-]=[NH_3]#, and #[HA]=[NH_4Cl]#.

For such a buffer #pH=pK_a+log_10{[[NH_3(aq)]]/[[NH_4Cl(aq)]]}#

And I also ASSUME, that the required #pH=8.90#

Now #pK_a=9.24# for ammonium ion.

And thus, substituting these values into the equation..........

#log_10{[[NH_3(aq)]]/[[NH_4Cl(aq)]]}=-0.34#

i.e. #([NH_3(aq)])/([NH_4Cl(aq)])=10^(-0.34)#,

i.e. #"moles of ammonium chloride"xx0.457="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*mol#

#"ammonium chloride"=(5.00xx10^-2*mol)/(0.457)=0.109*mol#

And thus we have to add............

#0.109*molxx53.49*g*mol^-1=5.85*g#

To the initial #250.0*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)}#

#=8.90# as required..........