Question

What is the Ka for an acid HA, if the equilibrium concentrations are [HA]=0.822 M, [H3O+]=1.15×10−4 M, and [A−]=1.15×10−4 M?

Report your answer in scientific notation.
Your answer should have three significant figures.

Answer 1

The acid ionization constant (Ka) for the weak acid HA is calculated as 1.61×10⁻⁷, using the given equilibrium concentrations of HA, H3O+, and A- in the Ka expression.

Step-by-step explanation:

The acid ionization constant (Ka) for a weak acid HA can be calculated using the equilibrium concentrations of the acid (HA), the hydronium ion (H3O+), and the conjugate base (A-). The Ka expression for the weak acid HA is given by the concentration of the products over the concentration of the reactants, which is Ka = [H3O+][A-] / [HA]. In this case, the equilibrium concentrations are [HA] = 0.822 M, [H3O+] = 1.15×10⁻⁴ M, and [A⁻] = 1.15×10⁻⁴ M.

Plugging the concentrations into the expression, we get:

Ka = (1.15×10⁻⁴ M × 1.15×10⁻⁴ M) / 0.822 M

Ka = 1.61×10⁻⁷ (to three significant figures)

The Ka for the acid HA is 1.61×10⁻⁷ in scientific notation with three significant figures.

Answer 2

[.822] that’s the answer