Question

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

Answer 1

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..........