Question

what is the ka of the acid ha given that a 1.20 m solution of the acid has a ph of 0.20? the equation described by the ka value is ha(aq) h2o(l)⇌a−(aq) h3o (aq)TWO SIGNIFICANT FIGURES PLEASE

Answer 1

The pH of a solution is related to the concentration of H3O+ ions by the equation:

pH = -log[H3O+]

Therefore, we can calculate the concentration of H3O+ ions in the solution as:

[H3O+] = 10^(-pH) = 10^(-0.20) = 0.63 M

Since the acid, HA, is a weak acid, it undergoes partial dissociation in water according to the equation:

HA(aq) + H2O(l) ⇌ A-(aq) + H3O+(aq)

where A- is the conjugate base of HA.

The equilibrium constant expression for this reaction is:

Ka = [A-][H3O+]/[HA]

At equilibrium, the concentration of HA that remains undissociated is equal to the initial concentration of the acid, since the dissociation is only partial.

Thus, we have:

[HA] = 1.20 M

[A-] = [H3O+] = 0.63 M

Substituting these values into the equilibrium constant expression, we get:

Ka = (0.63 M)^2 / (1.20 M) = 0.3315

Rounding to two significant figures, the Ka of the acid HA is 0.33.

Step-by-step explanation: