for the example 1: calculating the % dissociation, the part where the ICE table is used and you can use the quadratic formula to find concentration "x", the two answers I got for x was x= -0.01285M and x=0.01245M. You guys said the concentration I should have found is 0.0126M. I was trying to figure out which of the "x" is the correct one ( I assume since a negative concentration can not exist, the concentration has to be 0.01245M) and I have gone through my calculations a few times, and I don't know where I went wrong. Is it a rounding error?

What you've calculated using the quadratic formula is correct. As you rightly say, you can't have a negative concentration, so the viable answer is 0.01245 M. If you use the small x assumption, then it comes out as 0.012649 M. You would expect a small difference in the result depending on which method is used. So it's strange that the article says that either method will give x=0.0126 M. My guess is that the author of the article probably used the small x method and perhaps didn't check the quadratic formula result. However, the difference is so small that it's negligible,

In example 1, why is the formula for % dissociation [A-]/[HA]*100% and not [H3O+]/[HA]*100% or [H3O+][A-]/[HA]*100%? Is this a stupid question? Sorry, if it is.

It's not a stupid question. You can use either [A⁻]/[HA]₀× 100 % or [H₃O⁺]/[HA]₀× 100 % to calculate % dissociation of a weak acid. Your last formula is just Kₐ×100 %, so that one is wrong.

So all of these are happening in water. What if these reactions aren't happening in water? Is there a situation like that? Like in gas? or something? Not something necessary to think about?

This is an interesting area, but I don't know much about this – I certainly don't remember learning about this in 1st or 2nd year undergraduate chemistry. However, acid-base reactions definitely take place in solvents other than water and even in the gas phase. For non-aqueous solvents, the pKa values for an acid may be higher or lower than they are in water depending on the acid and the solvent. The wikipedia article on this is somewhat readable: https://en.wikipedia.org/wiki/Acid_dissociation_constant#Acidity_in_nonaqueous_solutions An example I found for a gas phase acid-base reaction: https://www.youtube.com/watch?v=60vtFe42sGs (NB: the demonstrator shows a deplorable lack of concern for modeling safe laboratory practices!)

Is it possible to find the percent dissociation of a weak base, or is it only applicable to weak acids?

You can find the percent ionization of a weak base. This is analogous to finding the percent dissociation of an acid, except you are interested in what percentage of the base became ionized by bonding to an H+ ion.

why are we making ICE tables here?

ICE tables are just a way of organizing data. You don't have to use them, but it often is one of the best ways to keep track of lots of different numbers.

Well i'm a 3rd grader and I want to learn this and isn't OH weak?

OH- is actually considered to be a strong base, as its conjugate acid, water (H2O), is a weak acid.

In the percent dissocation example above, and in the last step (step 4), why did we use the [HNO3] as 0.400 M rather than (0.400-x) which should be the more accurate concentration (after we found x=0.0126)?

Because that’s how percent ionisation is defined. It’s dissociated / initial.

Main content

Chemistry library

Course: Chemistry library > Unit 13

Lesson 2: Acid-base equilibria

Weak acid-base equilibria

Google Classroom

Weak acid and base ionization reactions and the related equilibrium constants, Ka and Kb. Relating Ka and Kb to pH, and calculating percent dissociation.

Key points:

For a generic monoprotic weak acid start text, H, A, end text with conjugate base start text, A, end text, start superscript, minus, end superscript, the equilibrium constant has the form:

K, start subscript, start text, a, end text, end subscript, equals, start fraction, open bracket, start text, H, end text, start subscript, 3, end subscript, start text, O, end text, start superscript, plus, end superscript, close bracket, open bracket, start text, A, end text, start superscript, minus, end superscript, close bracket, divided by, open bracket, start text, H, A, end text, close bracket, end fraction

The acid dissociation constant K, start subscript, start text, a, end text, end subscript quantifies the extent of dissociation of a weak acid. The larger the value of K, start subscript, start text, a, end text, end subscript, the stronger the acid, and vice versa.
For a generic weak base start text, B, end text with conjugate acid start text, B, H, end text, start superscript, plus, end superscript, the equilibrium constant has the form:

K, start subscript, start text, b, end text, end subscript, equals, start fraction, open bracket, start text, B, H, end text, start superscript, plus, end superscript, close bracket, open bracket, start text, O, H, end text, start superscript, minus, end superscript, close bracket, divided by, open bracket, start text, B, end text, close bracket, end fraction

The base dissociation constant (or base ionization constant) K, start subscript, start text, b, end text, end subscript quantifies the extent of ionization of a weak base. The larger the value of K, start subscript, start text, b, end text, end subscript, the stronger the base, and vice versa.

Strong vs. weak acids and bases

Strong acids and strong bases refer to species that completely dissociate to form ions in solution. By contrast, weak acids and bases ionize only partially, and the ionization reaction is reversible. Thus, weak acid and base solutions contain multiple charged and uncharged species in dynamic equilibrium.

In this article, we will discuss acid and base dissociation reactions and the related equilibrium constants: K, start subscript, start text, a, end text, end subscript, the acid dissociation constant, and K, start subscript, start text, b, end text, end subscript, the base dissociation constant.

Warm-up: Comparing acid strength and start text, p, H, end text

Problem 1: Weak vs. strong acids at the same concentration

We have two aqueous solutions: a 2, point, 0, start text, M, end text solution of hydrofluoric acid, start text, H, F, end text, left parenthesis, a, q, right parenthesis, and a 2, point, 0, start text, M, end text solution of hydrobromic acid, start text, H, B, r, end text, left parenthesis, a, q, right parenthesis. Which solution has the lower start text, p, H, end text?

(Choice A)
They have the same start text, p, H, end text
(Choice B)
start text, H, F, end text, left parenthesis, a, q, right parenthesis
(Choice C)
start text, H, B, r, end text, left parenthesis, a, q, right parenthesis

[I need help!!]

Problem 2: Weak vs. strong acids at different concentrations

This time we have a 2, point, 0, start text, M, end text solution of hydrofluoric acid, start text, H, F, end text, left parenthesis, a, q, right parenthesis, and a 1, point, 0, start text, M, end text solution of hydrobromic acid, start text, H, B, r, end text, left parenthesis, a, q, right parenthesis. Which solution has the lower start text, p, H, end text?

Assume we don't know the equilibrium constant for the dissociation of hydrofluoric acid.

(Choice A)
They have the same start text, p, H, end text
(Choice B)
start text, H, F, end text, left parenthesis, a, q, right parenthesis
(Choice C)
start text, H, B, r, end text, left parenthesis, a, q, right parenthesis
(Choice D)
We need more information!

[I need help!!]

Weak acids and the acid dissociation constant, K, start subscript, start text, a, end text, end subscript

Weak acids are acids that don't completely dissociate in solution. In other words, a weak acid is any acid that is not a strong acid.

The strength of a weak acid depends on how much it dissociates: the more it dissociates, the stronger the acid. In order to quantify the relative strengths of weak acids, we can look at the acid dissociation constant K, start subscript, start text, a, end text, end subscript, the equilibrium constant for the acid dissociation reaction.

For a generic monoprotic weak acid start text, H, A, end text, the dissociation reaction in water can be written as follows:

[What does "monoprotic" mean?]

start text, H, A, end text, left parenthesis, a, q, right parenthesis, plus, start text, H, end text, start subscript, 2, end subscript, start text, O, end text, left parenthesis, l, right parenthesis, \rightleftharpoons, start text, H, end text, start subscript, 3, end subscript, start text, O, end text, start superscript, plus, end superscript, left parenthesis, a, q, right parenthesis, plus, start text, A, end text, start superscript, minus, end superscript, left parenthesis, a, q, right parenthesis

Based on this reaction, we can write our expression for equilibrium constant K, start subscript, start text, a, end text, end subscript:

K, start subscript, start text, a, end text, end subscript, equals, start fraction, open bracket, start text, H, end text, start subscript, 3, end subscript, start text, O, end text, start superscript, plus, end superscript, close bracket, open bracket, start text, A, end text, start superscript, minus, end superscript, close bracket, divided by, open bracket, start text, H, A, end text, close bracket, end fraction

[I don't remember how to write equilibrium constant expressions...]

The equilibrium expression is a ratio of products to reactants. The more start text, H, A, end text dissociates into start text, H, end text, start superscript, plus, end superscript and the conjugate base start text, A, end text, start superscript, minus, end superscript, the stronger the acid, and the larger the value of K, start subscript, start text, a, end text, end subscript. Since start text, p, H, end text is related to open bracket, start text, H, end text, start subscript, 3, end subscript, start text, O, end text, start superscript, plus, end superscript, close bracket, the start text, p, H, end text of the solution will be a function of K, start subscript, start text, a, end text, end subscript as well as the concentration of the acid: the start text, p, H, end text decreases as the concentration of the acid and/or K, start subscript, start text, a, end text, end subscript increase.

Common weak acids

Carboxylic acids are a common functional group in organic weak acids, and they have the formula minus, start text, C, O, O, H, end text. Malic acid left parenthesis, start text, C, end text, start subscript, 4, end subscript, start text, H, end text, start subscript, 6, end subscript, start text, O, end text, start subscript, 5, end subscript, right parenthesis, an organic acid that contains two carboxylic acid groups, contributes to the tart flavor of apples and some other fruits. Since there are two carboxylic acid groups in the molecule, malic acid can potentially donate up to two protons.

The table below lists some more examples of weak acids and their K, start subscript, start text, a, end text, end subscript values.

Name	Formula	K, start subscript, start text, a, end text, end subscript, left parenthesis, 25, degrees, start text, C, end text, right parenthesis
Ammonium	start text, N, H, end text, start subscript, 4, end subscript, start superscript, plus, end superscript	5, point, 6, times, 10, start superscript, minus, 10, end superscript
Chlorous acid	start text, H, C, l, O, end text, start subscript, 2, end subscript	1, point, 2, times, 10, start superscript, minus, 2, end superscript
Hydrofluoric acid	start text, H, F, end text	7, point, 2, times, 10, start superscript, minus, 4, end superscript
Acetic acid	start text, C, H, end text, start subscript, 3, end subscript, start text, C, O, O, H, end text	1, point, 8, times, 10, start superscript, minus, 5, end superscript

Concept check: Based on the table above, which is a stronger acidminusacetic acid or hydrofluoric acid?

[Show answer]

Example 1: Calculating % dissociation of a weak acid

One way to quantify how much a weak acid has dissociated in solution is to calculate the percent dissociation. The percent dissociation for weak acid start text, H, A, end text can be calculated as follows:

$\text{% dissociation} =\dfrac{[\text A^-(aq)]}{[\text{HA}(aq)]} \times 100\%$

If nitrous acid left parenthesis, start text, H, N, O, end text, start subscript, 2, end subscript, right parenthesis has a K, start subscript, start text, a, end text, end subscript of 4, point, 0, times, 10, start superscript, minus, 4, end superscript at 25, degrees, start text, C, end text, what is the percent dissociation of nitrous acid in a 0, point, 400, start text, space, M, end text solution?

Let's go through this example step-by-step!

Step 1: Write the balanced acid dissociation reaction

First, let's write the balanced dissociation reaction of start text, H, N, O, end text, start subscript, 2, end subscript in water. Nitrous acid can donate a proton to water to form start text, N, O, end text, start subscript, 2, end subscript, start superscript, minus, end superscript, left parenthesis, a, q, right parenthesis:

start text, H, N, O, end text, start subscript, 2, end subscript, left parenthesis, a, q, right parenthesis, plus, start text, H, end text, start subscript, 2, end subscript, start text, O, end text, left parenthesis, l, right parenthesis, \rightleftharpoons, start text, H, end text, start subscript, 3, end subscript, start text, O, end text, start superscript, plus, end superscript, left parenthesis, a, q, right parenthesis, plus, start text, N, O, end text, start subscript, 2, end subscript, start superscript, minus, end superscript, left parenthesis, a, q, right parenthesis

Step 2: Write the expression for K, start subscript, start text, a, end text, end subscript

From the equation in Step 1, we can write the K, start subscript, start text, a, end text, end subscript expression for nitrous acid:

K, start subscript, start text, a, end text, end subscript, equals, start fraction, open bracket, start text, H, end text, start subscript, 3, end subscript, start text, O, end text, start superscript, plus, end superscript, close bracket, open bracket, start text, N, O, end text, start subscript, 2, end subscript, start superscript, minus, end superscript, close bracket, divided by, open bracket, start text, H, N, O, end text, start subscript, 2, end subscript, close bracket, end fraction, equals, 4, point, 0, times, 10, start superscript, minus, 4, end superscript

Step 3: Find open bracket, start text, H, end text, start superscript, plus, end superscript, close bracket and open bracket, start text, N, O, end text, start subscript, 2, end subscript, start superscript, minus, end superscript, close bracket at equilibrium

Next, we can use an start text, I, C, E, end text table to determine algebraic expressions for the equilibrium concentrations in our K, start subscript, start text, a, end text, end subscript expression:

	start text, H, N, O, end text, start subscript, 2, end subscript, left parenthesis, a, q, right parenthesis	start text, H, end text, start subscript, 3, end subscript, start text, O, end text, start superscript, plus, end superscript	start text, N, O, end text, start subscript, 2, end subscript, start superscript, minus, end superscript
Initial	0, point, 400, start text, M, end text	0	0
Change	minus, x	plus, x	plus, x
Equilibrium	0, point, 400, start text, M, end text, minus, x	x	x

Plugging the equilibrium concentrations into our K, start subscript, start text, a, end text, end subscript expression, we get:

K, start subscript, start text, a, end text, end subscript, equals, start fraction, left parenthesis, x, right parenthesis, left parenthesis, x, right parenthesis, divided by, left parenthesis, 0, point, 400, start text, M, end text, minus, x, right parenthesis, end fraction, equals, 4, point, 0, times, 10, start superscript, minus, 4, end superscript

Simplifying this expression, we get the following:

start fraction, x, squared, divided by, 0, point, 400, start text, M, end text, minus, x, end fraction, equals, 4, point, 0, times, 10, start superscript, minus, 4, end superscript

This is a quadratic equation that can be solved for x either by using the quadratic formula or an approximation method.

[I don't know how to solve for x.]

Either method will give x, equals, 0, point, 0126, start text, space, M, end text. Therefore, open bracket, start text, N, O, end text, start subscript, 2, end subscript, start superscript, minus, end superscript, close bracket, equals, open bracket, start text, H, end text, start subscript, 3, end subscript, start text, O, end text, start superscript, plus, end superscript, close bracket, equals, 0, point, 0126, start text, space, M, end text.

Step 4: Calculate percent dissociation

To calculate percent dissociation, we can use the equilibrium concentrations we found in Step 3:

$\begin{aligned}{\text{% dissociation}}&=\dfrac{[\text{NO}_2^-]}{[\text{HNO}_2]}\\ \\ &=\dfrac{0.0126\text{ M}}{0.400\text{ M}}\times100\%\\ \\ &=3.2\%\end{aligned}$

Therefore, 3, point, 2, percent of the start text, H, N, O, end text, start subscript, 2, end subscript in solution has dissociated into start text, H, end text, start superscript, plus, end superscript and start text, N, O, end text, start subscript, 2, end subscript, start superscript, minus, end superscript ions.

Weak bases and K, start subscript, start text, b, end text, end subscript

Let's now examine the base dissociation constant (also called the base ionization constant) K, start subscript, start text, b, end text, end subscript. We can start by writing the ionization reaction for a generic weak base start text, B, end text in water. In this reaction, the base accepts a proton from water to form hydroxide and the conjugate acid, start text, B, H, end text, start superscript, plus, end superscript:

start text, B, end text, left parenthesis, a, q, right parenthesis, plus, start text, H, end text, start subscript, 2, end subscript, start text, O, end text, left parenthesis, l, right parenthesis, \rightleftharpoons, start text, B, H, end text, start superscript, plus, end superscript, left parenthesis, a, q, right parenthesis, plus, start text, O, H, end text, start superscript, minus, end superscript, left parenthesis, a, q, right parenthesis

[Why is this called a "base dissociation constant" when the base doesn't dissociate?]

We can write the expression for equilibrium constant K, start subscript, start text, b, end text, end subscript as follows:

K, start subscript, start text, b, end text, end subscript, equals, start fraction, open bracket, start text, B, H, end text, start superscript, plus, end superscript, close bracket, open bracket, start text, O, H, end text, start superscript, minus, end superscript, close bracket, divided by, open bracket, start text, B, end text, close bracket, end fraction

From this ratio, we can see that the more the base ionizes to form start text, B, H, end text, start superscript, plus, end superscript, the stronger the base, and the larger the value of K, start subscript, start text, b, end text, end subscript. As such, the start text, p, H, end text of the solution will be a function of both the value of K, start subscript, start text, b, end text, end subscript as well as the concentration of the base.

Example 2: Calculating the start text, p, H, end text of a weak base solution

What is the start text, p, H, end text of a 1, point, 50, start text, space, M, end text solution of ammonia, start text, N, H, end text, start subscript, 3, end subscript? left parenthesis, K, start subscript, start text, b, end text, end subscript, equals, 1, point, 8, times, 10, start superscript, minus, 5, end superscript, right parenthesis

This example is an equilibrium problem with one extra step: finding start text, p, H, end text from open bracket, start text, O, H, end text, start superscript, minus, end superscript, close bracket. Let's go through the calculation step-by-step.

Step 1: Write the balanced ionization reaction

First, let's write out the base ionization reaction for ammonia. Ammonia will accept a proton from water to form ammonium, start text, N, H, end text, start subscript, 4, end subscript, start superscript, plus, end superscript:

start text, N, H, end text, start subscript, 3, end subscript, left parenthesis, a, q, right parenthesis, plus, start text, H, end text, start subscript, 2, end subscript, start text, O, end text, left parenthesis, l, right parenthesis, \rightleftharpoons, start text, N, H, end text, start subscript, 4, end subscript, start superscript, plus, end superscript, left parenthesis, a, q, right parenthesis, plus, start text, O, H, end text, start superscript, minus, end superscript, left parenthesis, a, q, right parenthesis

Step 2: Write the expression for K, start subscript, start text, b, end text, end subscript

From this balanced equation, we can write an expression for K, start subscript, start text, b, end text, end subscript:

K, start subscript, start text, b, end text, end subscript, equals, start fraction, open bracket, start text, N, H, end text, start subscript, 4, end subscript, start superscript, plus, end superscript, close bracket, open bracket, start text, O, H, end text, start superscript, minus, end superscript, close bracket, divided by, open bracket, start text, N, H, end text, start subscript, 3, end subscript, close bracket, end fraction, equals, 1, point, 8, times, 10, start superscript, minus, 5, end superscript

Step 3: Find open bracket, start text, N, H, end text, start subscript, 4, end subscript, start superscript, plus, end superscript, close bracket and open bracket, start text, O, H, end text, start superscript, minus, end superscript, close bracket at equilibrium

To determine the equilibrium concentrations, we use an start text, I, C, E, end text table:

	start text, N, H, end text, start subscript, 3, end subscript, left parenthesis, a, q, right parenthesis	start text, N, H, end text, start subscript, 4, end subscript, start superscript, plus, end superscript	start text, O, H, end text, start superscript, minus, end superscript
Initial	1, point, 50, start text, M, end text	0	0
Change	minus, x	plus, x	plus, x
Equilibrium	1, point, 50, start text, M, end text, minus, x	x	x

Plugging the equilibrium values into our K, start subscript, start text, b, end text, end subscript expression, we get the following:

K, start subscript, start text, b, end text, end subscript, equals, start fraction, left parenthesis, x, right parenthesis, left parenthesis, x, right parenthesis, divided by, 1, point, 50, start text, M, end text, minus, x, end fraction, equals, 1, point, 8, times, 10, start superscript, minus, 5, end superscript

Simplifying, we get:

start fraction, x, squared, divided by, 1, point, 50, start text, M, end text, minus, x, end fraction, equals, 1, point, 8, times, 10, start superscript, minus, 5, end superscript

This is a quadratic equation that can be solved by using the quadratic formula or an approximation method. Either method will yield the solution

x, equals, open bracket, start text, O, H, end text, start superscript, minus, end superscript, close bracket, equals, 5, point, 2, times, 10, start superscript, minus, 3, end superscript, start text, space, M, end text

Step 4: Find start text, p, H, end text from open bracket, start text, O, H, end text, start superscript, minus, end superscript, close bracket

Now that we know the concentration of hydroxide, we can calculate start text, p, O, H, end text:

\begin{aligned}\text{pOH}&=-\log[\text{OH}^-]\\ \\ &=-\log(5.2\times10^{-3})\\ \\ &=2.28\end{aligned}

Recall that at 25, degrees, start text, C, end text, start text, p, H, end text, plus, start text, p, O, H, end text, equals, 14. Rearranging this equation, we have:

start text, p, H, end text, equals, 14, minus, start text, p, O, H, end text

Plugging in our value for start text, p, O, H, end text, we get:

start text, p, H, end text, equals, 14, point, 00, minus, left parenthesis, 2, point, 28, right parenthesis, equals, 11, point, 72

Therefore, the start text, p, H, end text of the solution is 11.72.

Common weak bases

From soaps to household cleaners, weak bases are all around us. Amines, a neutral nitrogen with three bonds to other atoms (usually a carbon or hydrogen), are common functional groups in organic weak bases.

Amines act as bases because nitrogen's lone pair of electrons can accept an start text, H, end text, start superscript, plus, end superscript. Ammonia, start text, N, H, end text, start subscript, 3, end subscript is an example of an amine base. Pyridine, start text, C, end text, start subscript, 5, end subscript, start text, H, end text, start subscript, 5, end subscript, start text, N, end text, is another example of a nitrogen-containing base.

Summary

For a generic monoprotic weak acid start text, H, A, end text with conjugate base start text, A, end text, start superscript, minus, end superscript, the equilibrium constant has the form:

K, start subscript, start text, a, end text, end subscript, equals, start fraction, open bracket, start text, H, end text, start subscript, 3, end subscript, start text, O, end text, start superscript, plus, end superscript, close bracket, open bracket, start text, A, end text, start superscript, minus, end superscript, close bracket, divided by, open bracket, start text, H, A, end text, close bracket, end fraction

The acid dissociation constant K, start subscript, start text, a, end text, end subscript quantifies the extent of dissociation of a weak acid. The larger the value of K, start subscript, start text, a, end text, end subscript, the stronger the acid, and vice versa.
For a generic weak base start text, B, end text with conjugate acid start text, B, H, end text, start superscript, plus, end superscript, the equilibrium constant has the form:

K, start subscript, start text, b, end text, end subscript, equals, start fraction, open bracket, start text, B, H, end text, start superscript, plus, end superscript, close bracket, open bracket, start text, O, H, end text, start superscript, minus, end superscript, close bracket, divided by, open bracket, start text, B, end text, close bracket, end fraction

The base dissociation constant (or base ionization constant) K, start subscript, start text, b, end text, end subscript quantifies the extent of ionization of a weak base. The larger the value of K, start subscript, start text, b, end text, end subscript, the stronger the base, and vice versa.

[References and attributions]

Try it!

Problem 1: Finding K, start subscript, start text, b, end text, end subscript from start text, p, H, end text

A 1, point, 50, start text, space, M, end text solution of pyridine, start text, C, end text, start subscript, 5, end subscript, start text, H, end text, start subscript, 5, end subscript, start text, N, end text, has a start text, p, H, end text of 9, point, 70 at 25, degrees, start text, C, end text. What is the K, start subscript, start text, b, end text, end subscript of pyridine?

(Choice A)
1, point, 7, times, 10, start superscript, minus, 9, end superscript
(Choice B)
2, point, 6, times, 10, start superscript, minus, 20, end superscript
(Choice C)
3, point, 8, times, 10, start superscript, minus, 10, end superscript
(Choice D)
1, point, 8, times, 10, start superscript, minus, 5, end superscript

[Hint 1]

[Hint 2]

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p4q4storm
Posted 6 years ago. Direct link to p4q4storm's post “for the example 1: calcul...”
for the example 1: calculating the % dissociation, the part where the ICE table is used and you can use the quadratic formula to find concentration "x", the two answers I got for x was x= -0.01285M and x=0.01245M. You guys said the concentration I should have found is 0.0126M. I was trying to figure out which of the "x" is the correct one ( I assume since a negative concentration can not exist, the concentration has to be 0.01245M) and I have gone through my calculations a few times, and I don't know where I went wrong. Is it a rounding error?
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(7 votes)
Answer
- RogerP
  Posted 6 years ago. Direct link to RogerP's post “What you've calculated us...”
  What you've calculated using the quadratic formula is correct. As you rightly say, you can't have a negative concentration, so the viable answer is 0.01245 M.
  
  If you use the small x assumption, then it comes out as 0.012649 M. You would expect a small difference in the result depending on which method is used. So it's strange that the article says that either method will give x=0.0126 M.
  
  My guess is that the author of the article probably used the small x method and perhaps didn't check the quadratic formula result.
  
  However, the difference is so small that it's negligible,
  Comment on RogerP's post “What you've calculated us...”
  (17 votes)
Yuya Fujikawa
Posted 7 years ago. Direct link to Yuya Fujikawa's post “In example 1, why is the ...”
In example 1, why is the formula for % dissociation [A-]/[HA]*100% and not [H3O+]/[HA]*100% or [H3O+][A-]/[HA]*100%? Is this a stupid question? Sorry, if it is.
Button navigates to signup pageComment on Yuya Fujikawa's post “In example 1, why is the ...”
(5 votes)
Answer
- Ernest Zinck
  Posted 7 years ago. Direct link to Ernest Zinck's post “It's not a stupid questio...”
  It's not a stupid question.
  You can use either [A⁻]/[HA]₀× 100 % or [H₃O⁺]/[HA]₀× 100 % to calculate % dissociation of a weak acid.
  Your last formula is just Kₐ×100 %, so that one is wrong.
  Comment on Ernest Zinck's post “It's not a stupid questio...”
  (7 votes)
Hannah McGowen
Posted 7 years ago. Direct link to Hannah McGowen's post “Is it possible to find th...”
Is it possible to find the percent dissociation of a weak base, or is it only applicable to weak acids?
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(4 votes)
Answer
- yuki
  Posted 7 years ago. Direct link to yuki's post “You can find the percent ...”
  You can find the percent ionization of a weak base. This is analogous to finding the percent dissociation of an acid, except you are interested in what percentage of the base became ionized by bonding to an H+ ion.
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Yuya Fujikawa
Posted 6 years ago. Direct link to Yuya Fujikawa's post “So all of these are happe...”
So all of these are happening in water. What if these reactions aren't happening in water? Is there a situation like that? Like in gas? or something? Not something necessary to think about?
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(5 votes)
Answer
- tyersome
  Posted 6 years ago. Direct link to tyersome's post “This is an interesting ar...”
  This is an interesting area, but I don't know much about this – I certainly don't remember learning about this in 1st or 2nd year undergraduate chemistry. However, acid-base reactions definitely take place in solvents other than water and even in the gas phase.
  
  For non-aqueous solvents, the pKa values for an acid may be higher or lower than they are in water depending on the acid and the solvent. The wikipedia article on this is somewhat readable:
  https://en.wikipedia.org/wiki/Acid_dissociation_constant#Acidity_in_nonaqueous_solutions
  
  An example I found for a gas phase acid-base reaction:
  https://www.youtube.com/watch?v=60vtFe42sGs
  (NB: the demonstrator shows a deplorable lack of concern for modeling safe laboratory practices!)
  Comment on tyersome's post “This is an interesting ar...”
  (5 votes)
Romaa
Posted 6 years ago. Direct link to Romaa's post “why are we making ICE tab...”
why are we making ICE tables here?
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(3 votes)
Answer
- Jonathan Ziesmer
  Posted 6 years ago. Direct link to Jonathan Ziesmer's post “ICE tables are just a way...”
  ICE tables are just a way of organizing data. You don't have to use them, but it often is one of the best ways to keep track of lots of different numbers.
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  (4 votes)
Jadyn Newberry
Posted 6 years ago. Direct link to Jadyn Newberry's post “Well i'm a 3rd grader and...”
Well i'm a 3rd grader and I want to learn this and isn't OH weak?
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(2 votes)
Answer
- Katie Schleicher
  Posted 6 years ago. Direct link to Katie Schleicher's post “OH- is actually considere...”
  OH- is actually considered to be a strong base, as its conjugate acid, water (H2O), is a weak acid.
  Comment on Katie Schleicher's post “OH- is actually considere...”
  (5 votes)
Bibika
Posted 6 years ago. Direct link to Bibika's post “After reading the article...”
After reading the article I understood that ICE Table applies only to the weak acid and bases and not to the strong acid and bases. In case of the strong acid and base we can directly use the concentration of the compound given because it dissociates totally. I am correct right?
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(3 votes)
Answer
- Dulyana Apoorva
  Posted 4 months ago. Direct link to Dulyana Apoorva's post “I guess you are correct, ...”
  I guess you are correct, because, as strong acids and bases dissociate completely in an aqueous solution, it is safe to say that their concentrations can be used in calculations. ICE literally stands for Initial, Change and Equilibrium, so, while it IS true that we have an equilibrium in even strong acids and bases, I think the reaction is favored so strong in the direction of the forward reaction of dissociation, so, the effect of the reverse reaction is negligible. Hence, there is no need for ICE tables.
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  (1 vote)
mkiwan
Posted 3 years ago. Direct link to mkiwan's post “In the percent dissocatio...”
In the percent dissocation example above, and in the last step (step 4), why did we use the [HNO3] as 0.400 M rather than (0.400-x) which should be the more accurate concentration (after we found x=0.0126)?
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(2 votes)
Answer
- Ryan W
  Posted 3 years ago. Direct link to Ryan W's post “Because that’s how percen...”
  Because that’s how percent ionisation is defined. It’s dissociated / initial.
  Comment on Ryan W's post “Because that’s how percen...”
  (3 votes)
hammondkristen3
Posted 3 years ago. Direct link to hammondkristen3's post “Which is more dangerous: ...”
Which is more dangerous: a dilute strong acid or a concentrated weak base.
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(2 votes)
Answer
Jayom Raval
Posted 3 years ago. Direct link to Jayom Raval's post “In the ICE tables, is the...”
In the ICE tables, is the change always -x? If not, under what conditions would be higher (e.g. -2x or some other number)?
Thank you
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(2 votes)
Answer

Chemistry library

Course: Chemistry library > Unit 13

Key points:

Strong vs. weak acids and bases

Warm-up: Comparing acid strength and pH\text{pH}pHstart text, p, H, end text

Problem 1: Weak vs. strong acids at the same concentration

Problem 2: Weak vs. strong acids at different concentrations

Weak acids and the acid dissociation constant, KaK_\text{a}Ka​K, start subscript, start text, a, end text, end subscript

Common weak acids

Example 1: Calculating % dissociation of a weak acid

Step 1: Write the balanced acid dissociation reaction

Step 2: Write the expression for KaK_\text{a}Ka​K, start subscript, start text, a, end text, end subscript

Step 4: Calculate percent dissociation

Weak bases and KbK_\text{b}Kb​K, start subscript, start text, b, end text, end subscript

Example 2: Calculating the pH\text{pH}pHstart text, p, H, end text of a weak base solution

Step 1: Write the balanced ionization reaction

Step 2: Write the expression for KbK_\text{b}Kb​K, start subscript, start text, b, end text, end subscript

Step 4: Find pH\text{pH}pHstart text, p, H, end text from [OH−][\text{OH}^-][OH−]open bracket, start text, O, H, end text, start superscript, minus, end superscript, close bracket

Common weak bases

Summary

Try it!

Problem 1: Finding KbK_\text{b}Kb​K, start subscript, start text, b, end text, end subscript from pH\text{pH}pHstart text, p, H, end text

Want to join the conversation?

Warm-up: Comparing acid strength and start text, p, H, end text

Weak acids and the acid dissociation constant, K, start subscript, start text, a, end text, end subscript

Step 2: Write the expression for K, start subscript, start text, a, end text, end subscript

Weak bases and K, start subscript, start text, b, end text, end subscript

Example 2: Calculating the start text, p, H, end text of a weak base solution

Step 2: Write the expression for K, start subscript, start text, b, end text, end subscript

Step 4: Find start text, p, H, end text from open bracket, start text, O, H, end text, start superscript, minus, end superscript, close bracket

Problem 1: Finding K, start subscript, start text, b, end text, end subscript from start text, p, H, end text