I have to prepare a NH

_{3}/NH

_{4}^{+} buffer solution at pH = 10 with [NH

_{4}^{+}] = 0,05M

The starting solution are:

a) NH

_{3} solution, 2M

b) NH

_{4}^{+} salt 1:1 , 0.1M

So I think I should use the H.H.equation:

pH= pka + log ( NH

_{3}/NH

_{4}^{+} )

10 = 9.25 + log [NH

_{3}] - log (0.05)

[NH

_{3}] = 10^(-0.55) = 0,28M

SO I have 250ml of buffer solution at pH= 10 with:

[NH

_{3}]= 0,28 M

[NH

_{4}^{+} ] = 0,05 M

Now I can find the correspondents moles in those 250ml (0,250l) :

mol NH

_{3} = 0, 28M * 0,250l = 0,07mol

mol NH

_{4}^{+} = 0,05M * 0,250l = 0,0125 mol

Now I consider my NH3 and NH4 starting solutions in order to find the corresponding volume that cointains that number of moles I "need" :

V NH3 = 0,07mol / 2M = 0,035l

V NH4 = 0,0125 mol / 0,1 = 0,125l

SO I have to mix

0,035l of the starting NH3 solution with 0,125l of the starting NH4+ solution.

0,035l + 0,125l = 0,16l (V NH3 + V NH4+ added)

0,250 - 0,16 = 0,09 ml

SO I add ,at the end, 0,09ml of water in order to get 250ml

Is it correct?

THANKS