What is the pH of 0.2 M acetic acid ka equals 1.8 x 10-5?

Answer 1

What is the pH of 0.2 M acetic acid Ka equals 1.8 x 10-5? This is an equilibrium problem.

When HC2H3O dissolved in water, it forms H3O+1 ions and C2H3O2-1 ions. As the H3O+1 ions and C2H3O2-1 ions bounce around in the water, they reform water and HC2H3O2 molecules.

#1 HC2H3O2 + H2O = H3O+1 + C2H3O2-1

Ka is the dissociation constant. The dissociation constant is a measure of the number of number of H3O+1 and C2H3O2-1 ions per liter of solution compared to the original number of HC2H3O2 molecules per liter of solution. The brackets [ ] mean moles/liter, which is a measure of the concentration of the substance inside the brackets..

#2 Ka = [H3O+1] * [C2H3O2-1] ÷ [HC2H3O2], is the dissociation constant equation.

When one molecule of HC2H3O2 dissociates, in water, it forms one H3O+1 ion and one C2H3O2-1 ion.

A 0.2 molar solution of HC2H3O2 means 0.2 moles of HC2H3O2 was dissolved in enough water to make one liter of solution

#1 HC2H3O2 + H2O = H3O+1 + C2H3O2-1

#2 Ka = ( [H3O+1] * [C2H3O2-1] ) ÷ [HC2H3O2]

We do not include the concentration of water, because the number of water molecules that do not dissociate greatly exceeds the number that do dissociate. We know this because the Ka is very small.

Since we do not know how many HC2H3O2 dissociated, let x equal the moles/per liter of HC2H3O that dissociated forming x moles/per liter of H3O+1 ions and x moles/per liter of C2H3O2-1 ions. Now, substitute x for the [H3O+1] and. [C2H3O2-1].

#3 Ka = x * x ÷ [HC2H3O2]

#4 Ka = x^2 ÷ [HC2H3O2]

#5 x^2 = Ka ÷ [HC2H3O2]

#6 x = ( Ka ÷ [HC2H3O2] ) ^0.5

Now, substitute Ka = 1.8 x 10-5 for the dissociation, and 0.2 for [HC2H3O2].

Solve for the value of x.

(You can substitute these values of Ka and [HC2H3O2] into equation #3 and solve for x. I just wanted to show you how to derive the equation.)

This answer will be very small, around n x 10^-3.

x = [H3O+1] = about 10^-3.

pH = - log of the [H3O+1]

pH = - log x

Don't forget the negative sign.

This answer will be between +2 and +3

In case you do not know how to do pH problems, I will do an example.

What is the pH of 0.1 molar solution of HOBr? (Ka = 2.0 x 10^-9

Eq.#1 HOBr + H2O = H3O+1 + Br-1

Eq.#2 Ka = ( [H3O+1] * [Br-1] ) ÷ [HBr]

Eq.#3 Ka = x * x ÷ [HBr]

Eq.#4 Ka = x^2 ÷ [HBr]

Eq.#5 x^2 = Ka ÷ [HBr]

Now, substitute Ka = 2.0 x 10^-9 for the dissociation, and 0.1 for [HBr].(10^-1)

Solve for the value of x.

Eq.#5 x^2 = (2.0 x 10^-9) ÷ (10^-1)

Dividing powers, means subtracting exponents!! (-9) - (-1) = -8

Eq.#5 x^2 = (2.0 x 10^-8)

Remember, when you take (n * 10^-8)^0.5 = n^0.5 * (10 ^-8)^0.5

Remember, when you take the square root of 10^-8, you get 10^-4, because 10^-4 * 10^-4 = 10^-8

Eq.#6 x = (2.0 x 10^-8) ^0.5=

x = [H3O+1] = 1.414 * 10^-4

pH = - log of the [H3O+1]

log 10^x = exponent of 10 = x

Log of ( n * 10^-b) = (log n) + (-b)

pH = - log (1.414 * 10^-4 )

pH = (- log 1.414 + -log * 10^-4 ) = (-0.15) + (+4) = +3.85

Don't forget 2 negatives = a positive +4

This answer will be between +3 and +4