Answer:
111.95mL of HNO3 are needed to prepare the buffer
Step-by-step explanation:
We can solve this equation using H-H equation for bases:
pOH = pKb + log [HA+] / [A]
Where pOH is the pOH of the solution
pOH = 14 - pH = 14 - 8.970 = 5.03
pKb is the pKb of NH3 = 4.74
[HA+] could be taken as moles of NH4+
[A] as moles of NH3
The NH3 reacts with nitric acid, HNO3, as follows:
NH3 + HNO3 → NH4+ + NO3-
That means the moles of HNO3 added = X = Moles of NH4+ produced
And moles of NH3 are initial moles NH3 - X
Initial moles of NH3 are:
0.125L * (0.374mol/L) = 0.04675 moles NH3
Replacing in H-H equation:
pOH = pKb + log [HA+] / [A]
5.03 = 4.74 + log [X] / [0.04675-X]
0.29 = log [X] / [0.04675-X]
1.95 = [X] / [0.04675-X]
0.0912 - 1.95X = X
0.0912 = 2.95X
X = 0.0309 moles
We need to add 0.0309 moles of HNO3. From a solution that is 0.276M:
0.0309 moles of HNO3 * (1L / 0.276moles) = 0.11195L of HNO3 are needed
In mL:
111.95mL of HNO3 are needed to prepare the buffer