When we have PH = 9.45 so, we can get POH from:
PH = 14 - POH
POH = 14 - 9.45
= 4.55
So by using H-H equation, we can get [NH4+]:
POH = Pkb + ㏒ [B-/HB-]
when Kb = 1.8 x 10^-5
∴Pkb = -㏒(1.8 x 10^-5)
= 4.7
by substitution:
4.55 = 4.7 + ㏒[0.375]/[NH4]
∴[NH4] = 0.53 M
now we need to get the moles of NH4 at 0.25 L
moles NH4 = molarity * volume
= 0.53 M* 0.25L = 0.1325 moles
now, we can get the mass of NH4 when:
Mass NH4 = moles NH4 * molar mass
= 0.1325 moles * 53.49 g/mol
= 7 g
