Question

What is the concentration of hydronium ion, H3O + , in 0.10 M HClO at 25°C?

A. 3.0 × 10−8 M

B. 3.0 × 10−7 M

C. 5.5 × 10−5 M

D. 1.7 × 10−4 M

E. 0.10 M

Answer 1

What is the hydronium ion concentration and pH of a 0.10 M solution of hypochlorous acid, K_a = 3.5 x 10^-8?

In this case, the concentration is known to the 10^-2 place and the equilibrium constant is the magnitude of 10^-8. The difference in magnitude is 10⁶, therefore, x may easily be neglected. This simplifies the equation to:

Multiplying both sides by 0.10 yields:

x² = 3.5 x 10^-9

Taking the square root of both sides yields:

x = [H₃O⁺] = 5.9 x 10^-5 = [OCl^-] (also)

In tackling this problem, first note that the K_a is a small number, meaning hypochlorous acid is a weak acid. To begin the problem, write down the equilibrium involved:

HOCl + H₂O = H₃O⁺ + OCl^-

The equilibrium may be expressed mathematically by setting the K_a equal to the mass action expression:

Next, use the equilibrium to establish a table of initial conditions, change in equilibrium and final equilibrium conditions. Initially, the HOCl concentration is 0.10 M, the concentrations of H₃O⁺ and OCl^- are zero:

Note that water will not be included in the calculation since is the solvent.

In order for equilibrium to be established, some of HOCl must dissociate and form H₃O⁺ and OCl^-. Since it is not known how much will dissociate, we'll call the amount of HOCl lost -x and the amount of H₃O⁺ and OCl^- formed +x:

The above operation is justified by LeChatlier's Principle, which states: if a stress is places on an equilibrium, the equilibrium will shift in the direction which will relieve the stress. In this case, the stress is the lack of H₃O⁺ and OCl^- so the equilibrium will shift to the right to relieve the stress by forming some H₃O⁺ and OCl^-.

By summing the initial concentrations and the change in concentrations, you can obtain the amount of each species at equilibrium:

These quantities will be used in the mass action expression for the equilibrium of the acid, as shown below. The K_a for HOCl is 3.5 x 10^-8.

The solution for x becomes simplified because the x shown in bold can be neglected. This x can be neglected because it will be negligibly small compared to the concentration, 0.10 M. To determine whether x is negligible, compare the magnitude of the last decimal place of the concentration of the acid to the magnitude of the equilibrium constant. If the difference in magnitude is greater than 100, the x may be neglected.

This simplifies the equation to:

Multiplying both sides by 0.10 yields:

x² = 3.5 x 10^-9

Taking the square root of both sides yields:

x = [H₃O⁺] = 5.9 x 10^-5 = [OCl^-] (also)

Hence,Answer=C