The pH of a 0.1 M solution of an unknown weak acid is 3.7. What is the pKa of this acid?

Explanation:

Knowing the pH, you know the concentration of protons:

−log[H+]=pH=3.7

[H+]=10−3.7 M

Now, since the weak (monoprotic) acid dissociates into its conjugate base and a proton, the mols of protons are equimolar with the mols of conjugate base---the protons came FROM the weak acid, so the conjugate base that forms must be equimolar with the protons given out to the solvent.

HA⇌A−+H+

Hence, [A−]=[H+] in the same solution volume. Using the equilibrium constant expression, we get:

Ka=[H+]2eq[HA]eq

Don't forget that the HA form of HA had given away protons, so the mols of protons given away to generate A− is subtracted from the mols of (protons in) HA.

=[H+]2eq[HA]i−[H+]eq

=(10−3.7M)20.02M−10−3.7M

Ka=2.0105×10−6 M

here's the link to your question!

explanation:

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