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
anor277
Apr 25, 2017

Doubtless you mean ammonium chloride, NH4Cl. 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−]=[NH3], and [HA]=[NH4Cl].

For such a buffer pH=pKa+log10{[NH3(aq)][NH4Cl(aq)]}

And I also ASSUME, that the required pH=8.90

Now pKa=9.24 for ammonium ion.

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

log10{[NH3(aq)][NH4Cl(aq)]}=−0.34

i.e. [NH3(aq)][NH4Cl(aq)]=10−0.34,

i.e. moles of ammonium chloride×0.457=moles of ammonia, BECAUSE the volume was the same in each case.

moles of ammonia=0.250⋅L×0.200⋅mol⋅L−1=5.00×10−2⋅mol

ammonium chloride=5.00×10−2⋅mol0.457=0.109⋅mol

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

0.109⋅mol×53.49⋅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+log10⎧⎪⎨⎪⎩5.00×10−2⋅mol×10.2500⋅L5.85⋅g53.49⋅g⋅mol−1×10.2500⋅L⎫⎪⎬⎪⎭

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

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