Question

The ionization constant of a very weak acid, HA, is 4.1×10^−9. Calculate the equilibrium concentrations of H_3O^+, A^−, and HA in a 0.040M solution of the acid. Determine the concentrations of all species at equilibrium and the solution pH. [H_3O^+]=____mol/L [A^-]=_______mol/L [HA]=______mol/L pH= ​

Answer 1

To determine the equilibrium concentrations of H₃O⁺, A⁻, and HA for a weak acid with a given ionization constant, set up an ICE table and solve for x, which represents the change in concentrations. Then calculate pH using the negative logarithm of the H₃O⁺ concentration.

Step-by-step explanation:

To calculate the equilibrium concentrations of H3O+, A−, and HA in a 0.040 M solution of the very weak acid HA with an ionization constant (Ka) of 4.1×10−9, we would set up an ICE table for the dissociation of HA (HA → H3O+ + A−). Initially, the concentration of HA is 0.040 M, and the concentrations of H3O+ and A− are 0 M.

At equilibrium, the concentration changes are −x for HA and +x for both H3O+ and A−. The equilibrium expression is Ka = ([H3O+][A−]) / [HA]. Substituting the terms with x and solving for x using the quadratic formula or an approximation (since Ka is very small), we find the value of x, which equals the concentration of H3O+ and A−. The pH is then calculated by taking the negative logarithm of the concentration of H3O+.

The final equilibrium concentrations of HA will be approximately 0.040 M − x, because x is very small compared to the initial concentration of HA.