calculate the [H+] and [OH-] of a 3.66 x10^-7 M C2H3O2 solution

Answer 1

C₂H₃O₂ is assumed to be the acetate anion. It should be written with a negative charge as C₂H₃O₂^-

If this is indeed acetate, then we need to know the Kb for it, or the Ka for acetic acid.

Looking up Ka for acetic acid, I find it to be 1.76x10^-5

Kb = 1x10^-14 / 1.76x10^-5 = 5.68x10^-10

Write the hydrolysis reaction for C₂H₃O₂^-

C₂H₃O₂^- + H2O ==> C₂H₄O₂ + OH^-

Write the Kb expression

Kb = [C₂H₄O₂] [OH-] / [[C₂H₃O₂^-]

5.68x10^-10 = (x)(x) / 3.66x10^-7 - x

2.08x10^-16 - 5.68x10^-10x = x²

x² + 5.68x10^-10x - 2.08x10^-16 = 0

x = 1.4x10-8 M = [OH-]

Since this [OH^-] is so low, we must now also include the [OH^-] contributed from autohydrolysis of H₂O, which is 1x10^-7 M

Final [OH^- ] = 1.4x10^-8 + 1x10^-7 = 1.14x10^-7 M = [OH^-]

[H⁺] = 1x10^-14 / 1.14x10^-7

[H⁺] = 8.7710^-8 M

Answer 2

Is this 3.66x10^-7 the concentration of pH or pOH? If this is for pH, the pH is the negative log of the concentration.

Another way to express this is:

pH = -log[H+]

If you can find the pH you can use the 14 base scale and find the pOH and therefore that concentration because they both add up to 14, or pH + pOH= 14

pH= -log[3.66x10^-7] = 6.44

14 = 6.44 + pOH

pOH = 7.56

pOH = -log[OH]

7.56 = -log[OH]

10^-7.56 = 10^-log[OH]

or taking the antilog or inverse log of this number gives you:

10^-7.56 = 2.75 x 10^-8M for pOH. This would be the concentration of hydroxide ion or pOH.

If the first concentration provided was actually OH ion instead of H+ ion or Hydrogen concentration, then you would solve it in reverse, meaning you would use the first concentration to find pOH instead:

pOH = -log[OH]

pOH = -log[3.66x10^-7] =6.44 for the pOH

The pH would then have to be:

14 = 6.44 + pH

pH= 7.56

Then doing anti log or inverse function of log with this pH would give a concentration of:

pH = -log[H+]

7.56 = -log[H+]

10^-7.56 = 10^-log[H+] = 2.75x10^-8M for [H+].

So essentially the concentration amounts would be reversed.