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)
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

Explanation:

When the Kb is given, the Henderson-Hasselbalch equation can be used to calculate the pOH of a buffer solution:

pOH = pKb + log ([A⁻] / [HA]) = -log(Kb) + log ([BH+] / [B])

Here, moles can be used in place of the concentration since the pairs listed are both dissolved in 5L, which cancel due to the fraction in the logarithm.

a) pOH = -log(1.8 x 10⁻⁵) + log(1.5/1.0) = 4.92

pH = 14 - pOH = 14 - 4.92 = 9.08

b) pOH = -log(1.8 x 10⁻⁵) + log(1.0/1.5) = 4.57

pH = 14 - pOH = 14 - 4.57 = 9.43

c) pOH = -log(1.7 x 10⁻⁹) + log(1.5/1.0) =  8.95

pH = 14 - pOH = 14 - 8.95 = 5.05

d) pOH = -log(1.7 x 10⁻⁹) + log(1.0/1.5) =  8.59

pH = 14 - pOH = 14 -  = 5.41

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