Grade 12 buffer question?

NH3 is a weak base and NH4Cl is a salt of this weak base NH3 and strong acid HCl so its a basic buffer solution

and pOH of such solution is given by :

pOH = pKb + log [salt]/[base]

pOH = pKb + log [NH4Cl]/[NH3]

here the ratio [NH4Cl]/[NH3] represents = no.of moles of NH4Cl/no.of moles of NH3

now Ka = 5.59 X 10^-10

so pKa = -log Ka = -log (5.59 X 10^-10) = 9.253

as pKa + pKb = 14

so pKb = 14-pKa = 14-9.253 = 4.747

also pH = 9.120

and we know that pH + pOH = 14

so pOH = 14-pH = 14-9.12 = 4.88

putting in pOH expression...

4.88 = 4.747 + log NH4Cl]/[NH3]

4.88-4.747 = log [NH4Cl]/[NH3]

0.133 = log [NH4Cl]/[NH3]

taking anti log ...

[NH4Cl]/[NH3] = 10^0.133 = 1.358

no.of moles of NH4Cl/no.of moles of NH3 = 1.358

no.of moles of NH4Cl = no.of moles of NH3 X 1.358 = 0.7 X 1.358 = 0.951

so no. of moles of NH4Cl in the mixture is 0.951 ...

molecular mass of NH4Cl = 14 + 4 X 1 + 35.5 = 53.5 g/mole

so mass of NH4Cl recquired = no.of moles X molecular mass = 53.5 X 0.951 = 50.878 g

which is almost the same as your answer

feel free to ask any question

I'll tell you how to do it, but won't provide the work as I am too lazy. There is a shortcut method, specifically for buffer type questions. This will save you a whole lot of work so then you won't have to write out the stupid ICE table and balance the equation just to find the ratio of moles you need and convert. Just use the henderson hassalbachs equation: ph = pka + log moles base/ moles acid. Then once you solve for the moles, use stoichiometry to find the grams of NH4CL based on its molar mass. So... it would be set up like this... 9.120 = -log(5.59*10^-10) + log (moles nh4cl / 0.7). Then go on from there...