What is [H⁺] in a 0.330 M solution of acrylic acid, CH₂CHCOOH (Ka = 3.16 × 10⁻⁵)?

[H⁺] = 3.23x10⁻³ M

Explanation:

For a weak monoprotic acid, such as acrylic acid, we can calculate [H⁺] using the following formula if the solution concentration is significantly higher than the Ka:

Where C is the weak acid concentration.

We input the data:

And solve for [H⁺]:

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