Solve an equilibrium problem (using an ICE table) to...

Key Concepts

Acid Dissociation Equilibrium

This concept pertains to the reversible reaction in which a weak acid donates a proton to water, forming its conjugate base and hydronium ions. It is expressed with an equilibrium constant (Ka) that quantitatively describes how much of the acid dissociates in solution. Understanding acid dissociation equilibrium allows us to determine the concentrations of species at equilibrium, which is essential for calculating pH in weak acid systems.

ICE Table Method

An ICE table stands for Initial, Change, Equilibrium, and is a structured approach used to track the concentrations of reactants and products as a system reaches equilibrium. It simplifies the process of setting up equilibrium expressions and is particularly useful when dealing with weak acid or weak base reactions, where only a fraction of the acid or base ionizes in solution.

pH Calculation

pH is defined as the negative logarithm of the hydrogen ion concentration in a solution (pH = -log [H+]). Calculating pH in the context of equilibrium problems involves first finding the equilibrium concentration of H+ ions using equilibrium constants and ICE tables, then applying the logarithmic relationship. This fundamental concept is critical for understanding the acidity or basicity of a solution.

Buffer Solutions

A buffer solution consists of a weak acid and its conjugate base (or a weak base and its conjugate acid) and is capable of resisting changes in pH upon the addition of small amounts of acids or bases. The presence of both components enables the solution to moderate H+ concentration fluctuations, and the Henderson-Hasselbalch equation is often used to approximate the pH of such systems.

Common Ion Effect

The common ion effect describes the phenomenon where the addition of an ion already present in the equilibrium causes a shift in the position of the equilibrium, usually suppressing the ionization of a weak acid or weak base. This effect is crucial when evaluating the pH of solutions that contain mixtures of a weak acid and a salt providing its conjugate base, as it can significantly alter the degree of dissociation and thus the overall pH.

Transcript

00:01 So we need to solve an equilibrium problem using an ice table to calculate the ph of each solution.

00:05 The first one is 0 .15 molar hydrofluoric acid.

00:10 So in water, hydrofluoric acid will somewhat dissociate into hydrogen ions and fluoride ions.

00:17 So if you work out an ice chart, we're starting with 0 .15 molar hydrofluoric acid, and we don't have any of the products to begin with, some of that hydrofluic acid will dissolve.

00:30 To form hydrogen ions and fluoride ions, and they all have the same ratio, one to one to one, so they will increase and decrease by the same amounts.

00:41 So at equilibrium, you will get 0 .15 minus x, x and x.

00:47 And if we write our k -a expression, we have the hydrogen ions times the fluoride ions on the top, and our hydrofluoric acid on the bottom.

00:57 If you look up the ca value for hf, you get 6 .6 times 10 to the negative 4.

01:03 And i'm going to plug in my equilibrium values of x and x all over 0 .15 minus x.

01:13 If you solve for x, you get 0 .00963, and that's going to be equal to the hydrogen ion concentration x.

01:23 And from there, you can take the negative log of that number, which will give you the ph.

01:30 And you get that the ph is equal to 2 .02.

01:34 Now we need to solve an equilibrium problem using an ice table to calculate the ph of 0 .15 molar sodium fluoride.

01:41 For this part, this is completely soluble, so the sodium itself doesn't matter, but the fluoride does.

01:48 So the fluoride can react in water to form hydrofluoric acid and hydroxide ions.

01:58 And those hydroxide ions influence the ph of the solution.

02:03 So if we make an ice table, we know that the fluoride ions is going to be equal to 0 .15 because sodium fluoride is completely soluble, so it completely breaks down.

02:15 We don't include water because it's a liquid, and we don't have any hydrofluoric acid or hydroxide to start out with.

02:22 We know that because it's 1 to 1 to 1, the fluoride will decrease by x, and the acid and the hydroxide will increase by x.

02:32 So at equilibrium, it's 0 .15 minus x, x, and x.

02:36 Now this time we want to write a kb expression...

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