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

Course: Chemistry library > Unit 11

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 $HA$ ‍ with conjugate base $A^{-}$ ‍, the equilibrium constant has the form:

$K_{a} = \frac{[H_{3} O^{+}] [A^{-}]}{[HA]}$ ‍

The acid dissociation constant $K_{a}$ ‍ quantifies the extent of dissociation of a weak acid. The larger the value of $K_{a}$ ‍, the stronger the acid, and vice versa.
For a generic weak base $B$ ‍ with conjugate acid ${BH}^{+}$ ‍, the equilibrium constant has the form:

$K_{b} = \frac{[{BH}^{+}] [{OH}^{-}]}{[B]}$ ‍

The base dissociation constant (or base ionization constant) $K_{b}$ ‍ quantifies the extent of ionization of a weak base. The larger the value of $K_{b}$ ‍, 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_{a}

, the acid dissociation constant, and

K_{b}

, the base dissociation constant.

Warm-up: Comparing acid strength and $pH$ ‍

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

We have two aqueous solutions: a

2.0 M

solution of hydrofluoric acid,

HF (a q)

, and a

2.0 M

solution of hydrobromic acid,

HBr (a q)

. Which solution has the lower

pH

Answer:

HBr (a q)

has the lower

pH

We want to compare the relative amounts of

H^{+}

in solution to compare the

pH

values. Since the two acids have the same concentration, the acid that dissociates more in solution will have the lower

pH

HBr (a q)

is a strong acid that dissociates fully in solution:

$HBr (a q) \to {Br}^{-} (a q) + H^{+} (a q)$ ‍

In comparison,

HF (a q)

is a weak acid that only partially dissociates in solution:

$HF (a q) ⇋ F^{-} (a q) + H^{+} (a q)$ ‍

Since

HBr (a q)

is a stronger acid compared to

HF (a q)

, it will have a higher

[H^{+}]

and lower

pH

Problem 2: Weak vs. strong acids at different concentrations

This time we have a

2.0 M

solution of hydrofluoric acid,

HF (a q)

, and a

1.0 M

solution of hydrobromic acid,

HBr (a q)

. Which solution has the lower

pH

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

Answer: We need more information!

We want to compare the relative amounts of

H^{+}

(or

H_{3} O^{+}

) in solution to compare the

pH

values. However, this time the two acids do not have the same concentration:

HBr (a q)

is a strong acid that dissociates fully in solution, but we have a lower concentration of

HBr (a q)

compared to the weak acid

HF (a q)

In order to answer this question, we will need to calculate

[H^{+}]

for both of our solutions. This will require knowing the equilibrium constant for the dissociation of

HF (a q)

$HF (a q) ⇋ F^{-} (a q) + H^{+} (a q)$ ‍

K_{a} = \frac{[F^{-} (a q)] [H^{+} (a q)]}{[HF (a q)]}

Keep reading the article to learn more about solving this problem!

Weak acids and the acid dissociation constant, $K_{a}$ ‍

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_{a}

, the equilibrium constant for the acid dissociation reaction.

For a generic monoprotic weak acid

HA

, the dissociation reaction in water can be written as follows:

HA (a q) + H_{2} O (l) ⇌ H_{3} O^{+} (a q) + A^{-} (a q)

Based on this reaction, we can write our expression for equilibrium constant

K_{a}

K_{a} = \frac{[H_{3} O^{+}] [A^{-}]}{[HA]}

Here is a quick review of the rules for writing equilibrium constants:

Concentrations of products are multiplied together and placed in the numerator.
Concentrations of reactants are multiplied together and placed in the denominator.
All concentrations are raised to a power equal to the molar coefficient of each species from the balanced equation. (Therefore, before writing the equilibrium expression, we'll need to have a balanced equation!)
Only species in the aqueous and gaseous phases are included in the equilibrium expression. Pure solids and liquids are omitted.

For more details, check out Sal's video on reactions in equilibrium and the article on the equilibrium constant K.

The equilibrium expression is a ratio of products to reactants. The more

HA

dissociates into

H^{+}

and the conjugate base

A^{-}

, the stronger the acid, and the larger the value of

K_{a}

. Since

pH

is related to

[H_{3} O^{+}]

, the

pH

of the solution will be a function of

K_{a}

as well as the concentration of the acid: the

pH

decreases as the concentration of the acid and/or

K_{a}

increase.

Common weak acids

Carboxylic acids are a common functional group in organic weak acids, and they have the formula

- COOH

. Malic acid

(C_{4} H_{6} O_{5})

, 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_{a}

values.

Name	Formula	$K_{a} (25^{\circ} C)$ ‍
Ammonium	${NH}_{4}^{+}$ ‍	$5.6 \times 10^{- 10}$ ‍
Chlorous acid	${HClO}_{2}$ ‍	$1.2 \times 10^{- 2}$ ‍
Hydrofluoric acid	$HF$ ‍	$7.2 \times 10^{- 4}$ ‍
Acetic acid	${CH}_{3} COOH$ ‍	$1.8 \times 10^{- 5}$ ‍

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

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

HA

can be calculated as follows:

% dissociation = \frac{[A^{-} (a q)]}{[HA (a q)]} \times 100 %

If nitrous acid $({HNO}_{2})$ ‍ has a $K_{a}$ ‍ of $4.0 \times 10^{- 4}$ ‍ at $25^{\circ} C$ ‍, what is the percent dissociation of nitrous acid in a $0.400 M$ ‍ 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

{HNO}_{2}

in water. Nitrous acid can donate a proton to water to form

{NO}_{2}^{-} (a q)

{HNO}_{2} (a q) + H_{2} O (l) ⇌ H_{3} O^{+} (a q) + {NO}_{2}^{-} (a q)

Step 2: Write the expression for $K_{a}$ ‍

From the equation in Step 1, we can write the

K_{a}

expression for nitrous acid:

K_{a} = \frac{[H_{3} O^{+}] [{NO}_{2}^{-}]}{[{HNO}_{2}]} = 4.0 \times 10^{- 4}

Step 3: Find $[H^{+}]$ ‍ and $[{NO}_{2}^{-}]$ ‍ at equilibrium

Next, we can use an

ICE

table to determine algebraic expressions for the equilibrium concentrations in our

K_{a}

expression:

	${HNO}_{2} (a q)$ ‍	$H_{3} O^{+}$ ‍	${NO}_{2}^{-}$ ‍
Initial	$0.400 M$ ‍	$0$ ‍	$0$ ‍
Change	$- x$ ‍	$+ x$ ‍	$+ x$ ‍
Equilibrium	$0.400 M - x$ ‍	$x$ ‍	$x$ ‍

Plugging the equilibrium concentrations into our

K_{a}

expression, we get:

K_{a} = \frac{(x) (x)}{(0.400 M - x)} = 4.0 \times 10^{- 4}

Simplifying this expression, we get the following:

\frac{x^{2}}{0.400 M - x} = 4.0 \times 10^{- 4}

This is a quadratic equation that can be solved for

x

either by using the quadratic formula or an approximation method.

Either method will give

x = 0.0126 M

. Therefore,

[{NO}_{2}^{-}] = [H_{3} O^{+}] = 0.0126 M

Step 4: Calculate percent dissociation

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

\begin{aligned} % dissociation & = \frac{[{NO}_{2}^{-}]}{[{HNO}_{2}]} \\ = \frac{0.0126 M}{0.400 M} \times 100 % \\ = 3.2 % \end{aligned}

Therefore, $3.2 %$ ‍ of the ${HNO}_{2}$ ‍ in solution has dissociated into $H^{+}$ ‍ and ${NO}_{2}^{-}$ ‍ ions.

Weak bases and $K_{b}$ ‍

Let's now examine the base dissociation constant (also called the base ionization constant)

K_{b}

. We can start by writing the ionization reaction for a generic weak base

B

in water. In this reaction, the base accepts a proton from water to form hydroxide and the conjugate acid,

{BH}^{+}

B (a q) + H_{2} O (l) ⇌ {BH}^{+} (a q) + {OH}^{-} (a q)

We can write the expression for equilibrium constant

K_{b}

as follows:

K_{b} = \frac{[{BH}^{+}] [{OH}^{-}]}{[B]}

From this ratio, we can see that the more the base ionizes to form

{BH}^{+}

, the stronger the base, and the larger the value of

K_{b}

. As such, the

pH

of the solution will be a function of both the value of

K_{b}

as well as the concentration of the base.

Example 2: Calculating the $pH$ ‍ of a weak base solution

What is the

pH

of a

1.50 M

solution of ammonia,

{NH}_{3}

(K_{b} = 1.8 \times 10^{- 5})

This example is an equilibrium problem with one extra step: finding

pH

from

[{OH}^{-}]

. 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,

{NH}_{4}^{+}

{NH}_{3} (a q) + H_{2} O (l) ⇌ {NH}_{4}^{+} (a q) + {OH}^{-} (a q)

Step 2: Write the expression for $K_{b}$ ‍

From this balanced equation, we can write an expression for

K_{b}

K_{b} = \frac{[{NH}_{4}^{+}] [{OH}^{-}]}{[{NH}_{3}]} = 1.8 \times 10^{- 5}

Step 3: Find $[{NH}_{4}^{+}]$ ‍ and $[{OH}^{-}]$ ‍ at equilibrium

To determine the equilibrium concentrations, we use an

ICE

table:

	${NH}_{3} (a q)$ ‍	${NH}_{4}^{+}$ ‍	${OH}^{-}$ ‍
Initial	$1.50 M$ ‍	$0$ ‍	$0$ ‍
Change	$- x$ ‍	$+ x$ ‍	$+ x$ ‍
Equilibrium	$1.50 M - x$ ‍	$x$ ‍	$x$ ‍

Plugging the equilibrium values into our

K_{b}

expression, we get the following:

K_{b} = \frac{(x) (x)}{1.50 M - x} = 1.8 \times 10^{- 5}

Simplifying, we get:

\frac{x^{2}}{1.50 M - x} = 1.8 \times 10^{- 5}

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 = [{OH}^{-}] = 5.2 \times 10^{- 3} M

Step 4: Find $pH$ ‍ from $[{OH}^{-}]$ ‍

Now that we know the concentration of hydroxide, we can calculate

pOH

\begin{aligned} pOH & = - \log [{OH}^{-}] \\ = - \log (5.2 \times 10^{- 3}) \\ = 2.28 \end{aligned}

Recall that at

25^{\circ} C

pH + pOH = 14

. Rearranging this equation, we have:

pH = 14 - pOH

Plugging in our value for

pOH

, we get:

pH = 14.00 - (2.28) = 11.72

Therefore, the $pH$ ‍ 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

H^{+}

. Ammonia,

{NH}_{3}

is an example of an amine base. Pyridine,

C_{5} H_{5} N

, is another example of a nitrogen-containing base.

Summary

For a generic monoprotic weak acid $HA$ ‍ with conjugate base $A^{-}$ ‍, the equilibrium constant has the form:

$K_{a} = \frac{[H_{3} O^{+}] [A^{-}]}{[HA]}$ ‍

The acid dissociation constant $K_{a}$ ‍ quantifies the extent of dissociation of a weak acid. The larger the value of $K_{a}$ ‍, the stronger the acid, and vice versa.
For a generic weak base $B$ ‍ with conjugate acid ${BH}^{+}$ ‍, the equilibrium constant has the form:

$K_{b} = \frac{[{BH}^{+}] [{OH}^{-}]}{[B]}$ ‍

The base dissociation constant (or base ionization constant) $K_{b}$ ‍ quantifies the extent of ionization of a weak base. The larger the value of $K_{b}$ ‍, the stronger the base, and vice versa.

Try it!

Problem 1: Finding $K_{b}$ ‍ from $pH$ ‍

1.50 M

solution of pyridine,

C_{5} H_{5} N

, has a

pH

9.70

25^{\circ} C

. What is the

K_{b}

of pyridine?

We can make an

ICE

table for the ionization reaction to find the concentrations at equilibrium:

	$C_{5} H_{5} N (a q)$ ‍	$C_{5} H_{5} {NH}^{+}$ ‍	${OH}^{-} (a q)$ ‍
Initial	$1.50 M$ ‍	$0 M$ ‍	$0 M$ ‍
Change	$- x$ ‍	$+ x$ ‍	$+ x$ ‍
Equilibrium	$1.50 M - x$ ‍	$x$ ‍	$x$ ‍

Plugging our expressions for the equilibrium concentrations of each species into the

K_{b}

equation, we have:

K_{b} = \frac{(x) (x)}{(1.50 M - x)} = \frac{x^{2}}{1.50 M - x}

We can use

pH

to solve for

[{OH}^{-}]

, which will give us

x

. First we can use

pH

to calculate the

pOH

of the solution:

pOH = 14 - 9.70 = 4.30

We can then use

pOH

to solve for

[{OH}^{-}]

, which will give us

x

\begin{aligned} [{OH}^{-}] & = 10^{- pOH} \\ = 10^{- 4.30} \\ = 5.05 \times 10^{- 5} M \end{aligned}

Therefore,

x = 5.05 \times 10^{- 5} M

Finally, we can plug this value for

x

into our

K_{b}

expression to solve for

K_{b}

\begin{aligned} K_{b} & = \frac{x^{2}}{1.50 - x} \\ = \frac{(5.05 \times 10^{- 5})^{2}}{(1.50 - 5.05 \times 10^{- 5})} \\ = 1.7 \times 10^{- 9} \end{aligned}

Therefore, the $K_{b}$ ‍ of pyridine is $1.7 \times 10^{- 9}$ ‍.

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p4q4storm
Posted 7 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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(8 votes)
Answer
- RogerP
  Posted 7 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...”
  (18 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 8 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?
Button navigates to signup pageComment on Hannah McGowen's post “Is it possible to find th...”
(4 votes)
Answer
- yuki
  Posted 8 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.
  Button navigates to signup page
  (8 votes)
Yuya Fujikawa
Posted 7 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?
Button navigates to signup pageButton navigates to signup page
(5 votes)
Answer
- tyersome
  Posted 7 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 7 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 7 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.
  Button navigates to signup page
  (4 votes)
Jadyn Newberry
Posted 7 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 7 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 7 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)
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- Dulyana Apoorva
  Posted a year 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.
  Button navigates to signup page
  (1 vote)
mkiwan
Posted 4 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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- Ryan W
  Posted 4 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 4 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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Jayom Raval
Posted 4 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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Answer

Course: Chemistry library > Unit 11

Weak acid-base equilibria

Key points:

Strong vs. weak acids and bases

Warm-up: Comparing acid strength and $pH$ ‍

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, $K_{a}$ ‍

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 $K_{a}$ ‍

Step 3: Find $[H^{+}]$ ‍ and $[{NO}_{2}^{-}]$ ‍ at equilibrium

Step 4: Calculate percent dissociation

Weak bases and $K_{b}$ ‍

Example 2: Calculating the $pH$ ‍ of a weak base solution

Step 1: Write the balanced ionization reaction

Step 2: Write the expression for $K_{b}$ ‍

Step 3: Find $[{NH}_{4}^{+}]$ ‍ and $[{OH}^{-}]$ ‍ at equilibrium

Step 4: Find $pH$ ‍ from $[{OH}^{-}]$ ‍

Common weak bases

Summary

Attributions

Additional References

Try it!

Problem 1: Finding $K_{b}$ ‍ from $pH$ ‍

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Chemistry library

Course: Chemistry library > Unit 11

Key points:

Strong vs. weak acids and bases

Warm-up: Comparing acid strength and pH‍

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, Ka‍

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 Ka‍

Step 3: Find [H+]‍ and [NO2−]‍ at equilibrium

Step 4: Calculate percent dissociation

Weak bases and Kb‍

Example 2: Calculating the pH‍ of a weak base solution

Step 1: Write the balanced ionization reaction

Step 2: Write the expression for Kb‍

Step 3: Find [NH4+]‍ and [OH−]‍ at equilibrium

Step 4: Find pH‍ from [OH−]‍

Common weak bases

Summary

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Problem 1: Finding Kb‍ from pH‍

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Warm-up: Comparing acid strength and $pH$ ‍

Weak acids and the acid dissociation constant, $K_{a}$ ‍

Step 2: Write the expression for $K_{a}$ ‍

Step 3: Find $[H^{+}]$ ‍ and $[{NO}_{2}^{-}]$ ‍ at equilibrium

Weak bases and $K_{b}$ ‍

Example 2: Calculating the $pH$ ‍ of a weak base solution

Step 2: Write the expression for $K_{b}$ ‍

Step 3: Find $[{NH}_{4}^{+}]$ ‍ and $[{OH}^{-}]$ ‍ at equilibrium

Step 4: Find $pH$ ‍ from $[{OH}^{-}]$ ‍

Problem 1: Finding $K_{b}$ ‍ from $pH$ ‍