What combination of substances will give a buffered solution that has a pH of 5.05? (Assume each pair of substances is dissolved in 5.0 L of water.) (Kb for NH3 = 1.8 ´ 10–5; Kb for C5H5N = 1.7 ´ 10–9) Select one: a. 1.0 mole NH3 and 1.5 mole NH4Cl b. 1.5 mole NH3 and 1.0 mole NH4Cl c. 1.0 mole C5H5N and 1.5 mole C5H5NHCl d. 1.5 mole C5H5N and 1.0 mole C5H5NHCl e. none of these

c. 1.0 mole C5H5N and 1.5 mole C5H5NHCl

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]

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]

Right solution is:

question: ray weighs 100 kilograms, his wrestling partner, bobby, weighs 50 kilograms. if ray moves at 4 m s , how fast will bobby have to run at him to apply an equal amount of kinetic energy?

answer: 8 m/s he needs to run twice as fast to keep up so he has some work to do

explanation:   if the mass has units of kilograms and the velocity of meters per second, the kinetic energy has units of kilograms-meters squared per second squared.

question answered by

(jacemorris04)