Answer:
c. 1.0 mole C5H5N and 1.5 mole C5H5NHCl
Step-by-step explanation:
We can determine pH of a buffer using H-H equation:
pH = pKa + log [A⁻] / [HA]
Where [A⁻] is concentration of conjugate base and [HA] concentration of weak acid. These concentrations can be computed as moles of each species.
We need to determine pKa of both NH₃ and C₅H₅N buffers, thus:
pKb = -log Kb
NH₃ pKb = -log 1.8x10⁻⁵ = 4.74
C₅H₅N pKb = -log 1.7x10⁻⁹ = 8.77
And pKa = 14 - pKb:
NH₃ pKa = 14 - 4.74 = 9.26
C₅H₅N pKb = 14 - 8.77 = 5.23
A buffer works only under pH's between pKa-1 and pKa + 1. As pKa NH₃ buffer is 9.23 is not possible to produce a buffer with pH 5.05 for this system.
Thus, we only will compute the buffers made with C₅H₅N:
c. 1.0 mole C5H5N (Weak base) and 1.5 mole C5H5NHCl (Conjugate acid)
pH = pKa + log [A⁻] / [HA]
pH = 5.23+ log [1.0 moles] / 1.5 moles]
pH = 5.05
d. 1.5 mole C5H5N and 1.0 mole C5H5NHCl
pH = pKa + log [A⁻] / [HA]
pH = 5.23+ log [1.5 moles] / 1.0 moles]
pH = 5.41
Right solution is:
c. 1.0 mole C5H5N and 1.5 mole C5H5NHCl