A 1 liter solution contains 0.443 M acetic acid and 0.332 M sodium acetate. Addition of 0.365 moles of perchloric acid will: (Assume that the volume does not change upon the addition of perchloric acid.) Raise the pH slightly Lower the pH slightly Raise the pH by several units Lower the pH by several units Not change the pH Exceed the buffer capacity

Lower the pH several units

Explanation:

Typically, a buffer system would minimize changes in pH values with additions of strong acid or base. However, one should realize buffer systems do have limits. The limits are dictated by the amount of common ion in the buffer solution mix. The chemistry of the buffer and common ion functions to remove the excess strong acid or base added. However, if more strong acid or strong base is added that exceeds the capacity of the buffer solution to remove the an excess acid or base solution would result and pH values would change drastically.

In this problem, perchloric acid (HClO₄) is a strong acid delivering hydronium ion in an amount equal to the amount of perchloric acid noted in the problem.

Since the concentration of perchloric acid (HClO₄) and H⁺ ion exceeds the concentration of available acetate ion in the buffer solution, the buffer effect would be exhausted before removing all of the HClO₄. The remaining HClO₄ would then function to drastically reduce the pH of the remaining solution several units before stabilizing.

The 0.443M HOAc/0.332M NaOAc buffer before addition of the HClO₄ has a pH of 4.62.  Adding 0.365M in HClO₄ would cause the HOAc equilibrium to shift left because of the H⁺ overload. This shift would continue until all of the strong acid was removed or all of the conjugate base is exhausted by converting back to HOAc. That is, adding 0.365M HClO₄ removes all 0.332M OAc⁻ leaving 0.365M - 0.332M = 0.033M excess HClO₄. The pH of a 0.033M HClO₄ solution => pH = -logs[H⁺] = -log(0.033) = 1.48.

=> compare pH before adding acid = 4.62 to pH after adding acid = 1.48.

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

solute : a solute is a substance that is dissolved in another substance is known as a solvent. it is mixed with a special mixture know as homogeneous mixture.

solvent: a solvent is a substance that dissolved a solute, which results in a solution is known as solvent.