An analytical chemist is titrating 111.0 mL of a 0.3700 M solution of aniline (C6H5NH2) with a 0.3500 M solution of HNO3. The pK_b of aniline is 9.37. Calculate the pH of the base solution after the chemist has added 79.1 mL of the HNO_3 solution to it.

The answer is "4.31"

Explanation:

aniline millimoles

added  millimoles

millimoles aniline left

millimoles salt formed

total volume

first let us lay out the formula for potential

and kinetic energy:

pe = mgh

ke = (mv^2)/2

where: m=mass, v=velocity, g=gravitational

acceleration, h=height

calculating for mass using the known values of

pe:

340j = m (9.81 m/s^2) (50ft) (1m/3.28ft)

m = 2.27 kg

the law of conservation of energy states that

energy is neither created nor destroyed. therefore, the change in kinetic

energy from 50 ft to 35 ft would be equal to the change in potential energy

from 50 ft to 35 ft. however it would be opposite in signs since one is losing

while other is gaining.

ke(35ft) – ke(50ft) = - [ pe(35ft) – pe(50ft]

ke(50ft) = 0 since the brick is initially at rest

ke(35ft) = 340j - 2.27kg (9.81 m/s^2) (35ft)

(1m/3.28 ft)

ke(35ft)

= 102.38 j