Calculate the pH of a 0.393 M aqueous solution of phenol (a weak acid) (C6H5OH, Ka = 1.0×10-10) and the equilibrium concentrations of the weak acid and its conjugate base.

See explanation below

Explanation:

First to all, we need to write the general reaction for this.

The phenol is C₆H₅OH:

C₆H₅OH + H₂O > C₆H₅O⁻ + H₃O⁺

Now, to get the equilibrium concentrations, we need to write the ICE chart for this reaction:

       C₆H₅OH + H₂O > C₆H₅O⁻ + H₃O⁺

i)        0.393                               0            0

c)           -x                                 +x          +x

e)      0.393-x                              x            x

Now, let's write the equilibrium expression:

Ka = [C₆H₅O⁻] [H₃O⁺] / [C₆H₅OH]

Replacing the above values:

1x10⁻¹⁰ = x² / 0.393-x

However Ka is a very small value, so we can assume that the substration between 0.393-x would be very small, so we can round this to 0.393 only:

1x10⁻¹⁰ = x²/0.393

Solving for x:

x = √1x10⁻¹⁰ * 0.393

x = 6.27x10⁻⁶ M

This would be the equilibrium concentration of the conjugate base, the weak acid would be practically the same.

[C₆H₅OH] = 0.393 - 6.27x10⁻⁶ = 0.39399373 M

Finally for the pH:

pH = -log(6.27x10⁻⁶)

pH = 5.2

the tides are influenced by both the moon and the sun, it's easy to see that when the sun lines up with the moon and the earth, as during a new moon or full moon, the tidal effect is increased. these are known as spring tides, named not for the season, but for the fact that the water "springs" higher than normal. on the other hand, if the sun and the moon are 90 degrees apart in relation to an observer on earth as during the first quarter moon or third quarter moon, then high tides are not as high as they normally would be. this is because despite its greater distance, the sun's mass allows it to exert enough gravitational force on the oceans that it can negate some of the effects of the moon's pull. this phenomenon of lower high tides is called a neap tide. if you need more information on this topic visit moonconnection.com.