Weak acids and bases $K_a$ for weak acids (A-level only)

What you'll learn

What “weak” means for acids and bases in aqueous solution.
How to write a $K_a$ expression for a weak acid.
How to calculate pH from $K_a$ and concentration, and calculate $K_a$ from pH.
How to convert between $K_a$ and $pK_a$ , including the half-neutralisation method.

The prerequisites: acids, bases and pH

A Brønsted–Lowry acid is a proton donor. A Brønsted–Lowry base is a proton acceptor. A proton is a hydrogen ion, written as $H+\text{H}^+$ .

In aqueous solution, $H+\text{H}^+$ does not really float around alone — it bonds to water to form $H3O+\text{H}_3\text{O}^+$ . At A-Level, we usually write $H+\text{H}^+$ as a shorthand.

Definition

pH is a logarithmic measure of hydrogen ion concentration:

pH=−log⁡10[H+]\text{pH}=-\log_{10}[\text{H}^+]

In calculations, use the numerical value of $[H+][\text{H}^+]$ in mol dm⁻³.

Because pH is logarithmic, a change of 1 pH unit means a factor of 10 change in $[H+][\text{H}^+]$ .

Example

Converting pH into hydrogen ion concentration

A solution has pH 3.25. Find $[H+][\text{H}^+]$ .

Rearrange the pH equation by reversing the log:
$[H+]=10−pH[\text{H}^+]=10^{-\text{pH}}$
Substitute the pH value:
$[H+]=10−3.25[\text{H}^+]=10^{-3.25}$
Evaluate and include units:
$dm−3[\text{H}^+]=5.62 \times 10^{-4}\ \text{mol dm}^{-3}$

Strong, weak, concentrated and dilute

Dissociation means a substance separates into ions in solution. An aqueous solution is a solution in water.

A strong acid dissociates completely in water. For example, hydrochloric acid effectively produces one mole of $H+\text{H}^+$ for every mole of HCl dissolved.

A weak acid dissociates only slightly in water. Most acid particles remain as undissociated molecules.

Definition

Weak acid

A weak acid is an acid that only partially dissociates in aqueous solution, setting up a reversible equilibrium between the acid and its ions.

A weak base also forms ions only slightly in aqueous solution. For example, ammonia reacts only partially with water:

NH3(aq)+H2O(l)⇌NH4+(aq)+OH−(aq)\text{NH}_3\text{(aq)}+\text{H}_2\text{O(l)} \rightleftharpoons \text{NH}_4^+\text{(aq)}+\text{OH}^-\text{(aq)}

This section focuses mainly on $K_a$ , the equilibrium constant for weak acids.

Common Mistake

Weak does not mean dilute

Weak describes the extent of dissociation. Dilute describes low concentration. You can have a concentrated weak acid, or a dilute strong acid.

The weak acid equilibrium

Most A-Level weak acid calculations use a general weak monoprotic acid, HA. Monoprotic means each acid molecule can donate one proton.

The equilibrium is:

HA(aq)⇌H+(aq)+A−(aq)\text{HA(aq)} \rightleftharpoons \text{H}^+\text{(aq)}+\text{A}^-\text{(aq)}

The ion $A−\text{A}^-$ is the conjugate base of HA. A conjugate base is what remains after an acid has donated a proton.

An equilibrium is a dynamic situation where the forward and reverse reactions continue, but the concentrations of substances stay constant.

Key Idea

Weak acids are equilibrium systems

For a weak acid, HA, both undissociated acid molecules and ions are present at equilibrium. You must use equilibrium concentrations in $K_a$ expressions.

The acid dissociation constant, $K_a$

The acid dissociation constant, $K_a$ , measures how far a weak acid dissociates in water.

For:

HA(aq)⇌H+(aq)+A−(aq)\text{HA(aq)} \rightleftharpoons \text{H}^+\text{(aq)}+\text{A}^-\text{(aq)}

the expression is:

Ka=[H+][A−][HA]K_a=\frac{[\text{H}^+][\text{A}^-]}{[\text{HA}]}

Square brackets mean concentration in mol dm⁻³ at equilibrium.

For this type of expression, the units of $K_a$ are:

dm−3\frac{(\text{mol dm}^{-3})(\text{mol dm}^{-3})}{\text{mol dm}^{-3}}=\text{mol dm}^{-3}

Example

Constructing a Ka expression

Write the $K_a$ expression for ethanoic acid.

Write the dissociation equilibrium:
$CH3COOH(aq)⇌H+(aq)+CH3COO−(aq)\text{CH}_3\text{COOH(aq)} \rightleftharpoons \text{H}^+\text{(aq)}+\text{CH}_3\text{COO}^-\text{(aq)}$
Put the product concentrations on the top and the undissociated acid concentration on the bottom:
$Ka=[H+][CH3COO−][CH3COOH]K_a=\frac{[\text{H}^+][\text{CH}_3\text{COO}^-]}{[\text{CH}_3\text{COOH}]}$
Work out the units from the concentration terms:
$dm−3K_a=\text{mol dm}^{-3}$

Calculating the pH of a weak acid

Suppose the initial concentration of a weak acid HA is $c$ mol dm⁻³.

At equilibrium, let the concentration that dissociates be $x$ mol dm⁻³.

So:

$[H+]=x[\text{H}^+]=x$
$[A−]=x[\text{A}^-]=x$
$[HA]=c−x[\text{HA}]=c-x$

The exact expression is:

Ka=x2c−xK_a=\frac{x^2}{c-x}

For a weak acid, only a small amount dissociates, so $x$ is usually much smaller than $c$ . This lets you use the approximation:

\approx c

So:

Ka≈x2cK_a \approx \frac{x^2}{c}

and therefore:

[H+]≈Kac[\text{H}^+] \approx \sqrt{K_a c}

Key Idea

The square-root shortcut

For a weak monoprotic acid of concentration $c$ , if dissociation is small:

[H+]≈Kac[\text{H}^+] \approx \sqrt{K_a c}

Then use $pH=−log⁡10[H+]\text{pH}=-\log_{10}[\text{H}^+]$ .

Example

Calculating pH from Ka and concentration

Ethanoic acid has $dm−3K_a=1.74 \times 10^{-5}\ \text{mol dm}^{-3}$ . Calculate the pH of 0.100 mol dm⁻³ ethanoic acid.

Use the weak acid approximation:
$[H+]≈Kac[\text{H}^+] \approx \sqrt{K_a c}$
Substitute the values:
$[H+]≈(1.74×10−5)(0.100)[\text{H}^+] \approx \sqrt{(1.74 \times 10^{-5})(0.100)}$
Calculate the hydrogen ion concentration:
$dm−3[\text{H}^+] \approx \sqrt{1.74 \times 10^{-6}}=1.32 \times 10^{-3}\ \text{mol dm}^{-3}$
Convert to pH:
$pH=−log⁡10(1.32×10−3)=2.88\text{pH}=-\log_{10}(1.32 \times 10^{-3})=2.88$
Check the approximation is sensible:
$1.32×10−30.100×100=1.32%\frac{1.32 \times 10^{-3}}{0.100} \times 100=1.32\%$
Only a small percentage dissociated, so the approximation is reasonable.

Common Mistake

Using the starting concentration as [H+]

For a weak acid, the initial acid concentration $c$ is not the hydrogen ion concentration. Only a small fraction dissociates, so $[H+][\text{H}^+]$ must be found using $K_a$ .

Common Mistake

When the shortcut may not work

If the acid is not very weak, or the solution is extremely dilute, $\approx c$ may be a poor approximation. Then use the exact expression $Ka=x2c−xK_a=\frac{x^2}{c-x}$ instead.

Calculating $K_a$ from pH

If you know the pH and the initial concentration of a weak acid, you can work backwards.

First convert pH into $[H+][\text{H}^+]$ . For a weak monoprotic acid:

[A−]=[H+][\text{A}^-]=[\text{H}^+]

Then substitute into:

Ka=[H+][A−][HA]K_a=\frac{[\text{H}^+][\text{A}^-]}{[\text{HA}]}

Example

Calculating Ka from pH

A 0.0500 mol dm⁻³ weak monoprotic acid has pH 3.10. Calculate $K_a$ .

Convert pH to hydrogen ion concentration:
$dm−3[\text{H}^+]=10^{-3.10}=7.94 \times 10^{-4}\ \text{mol dm}^{-3}$
Use the weak acid relationships:
$dm−3[\text{A}^-]=7.94 \times 10^{-4}\ \text{mol dm}^{-3}$
and, using the small dissociation approximation:
$dm−3[\text{HA}] \approx 0.0500\ \text{mol dm}^{-3}$
Substitute into the $K_a$ expression:
$Ka=(7.94×10−4)(7.94×10−4)0.0500K_a=\frac{(7.94 \times 10^{-4})(7.94 \times 10^{-4})}{0.0500}$
Calculate:
$dm−3K_a=1.26 \times 10^{-5}\ \text{mol dm}^{-3}$

$pK_a$

$pK_a$ is another way of expressing acid strength. It is especially useful because many $K_a$ values are very small numbers written in standard form.

Definition

pKa

$pK_a$ is defined as:

pK_a=-\log_{10}K_a

It has no units.

To convert back:

K_a=10^{-pK_a}

A larger $K_a$ means a stronger weak acid. A smaller $pK_a$ also means a stronger weak acid.

Tip

Comparing acid strength

Think: large $K_a$ , low $pK_a$ , strong acid.

Example

Converting between Ka and pKa

Convert between $K_a$ and $pK_a$ .

For $dm−3K_a=6.3 \times 10^{-5}\ \text{mol dm}^{-3}$ :
$pKa=−log⁡10(6.3×10−5)=4.20pK_a=-\log_{10}(6.3 \times 10^{-5})=4.20$
For $pK_a=3.50$ :
$dm−3K_a=10^{-3.50}=3.16 \times 10^{-4}\ \text{mol dm}^{-3}$
Compare the two acids: the acid with $pK_a=3.50$ is stronger because it has the larger $K_a$ .

Finding $K_a$ from half-neutralisation

The half-neutralisation point is the point in a titration where exactly half the original acid has been neutralised by the added base.

For a weak acid HA titrated with a strong alkali such as NaOH:

HA+OH−→A−+H2O\text{HA}+\text{OH}^- \to \text{A}^-+\text{H}_2\text{O}

At half-neutralisation, half of the original HA has been converted into $A−\text{A}^-$ . Therefore:

[HA]=[A−][\text{HA}]=[\text{A}^-]

Substitute this into the $K_a$ expression:

Ka=[H+][A−][HA]K_a=\frac{[\text{H}^+][\text{A}^-]}{[\text{HA}]}

Since $[A−]=[HA][\text{A}^-]=[\text{HA}]$ , they cancel:

Ka=[H+]K_a=[\text{H}^+]

So:

pH=pKa\text{pH}=pK_a

This is why measuring the pH at half-neutralisation is a practical method for finding $K_a$ of a weak acid.

Weak acid strong base titration curve showing half-neutralisation where pH equals pKa

Example

Using half-neutralisation to find Ka

A weak monoprotic acid is titrated with NaOH. The equivalence point occurs after 30.0 cm³ of NaOH is added. At 15.0 cm³, the measured pH is 4.76. Calculate $K_a$ .

Identify the half-neutralisation volume:
$cm3\frac{30.0}{2}=15.0\ \text{cm}^3$
At half-neutralisation, use the relationship:
$pH=pKa\text{pH}=pK_a$
so:
$pK_a=4.76$
Convert $pK_a$ into $K_a$ :
$K_a=10^{-4.76}$
Calculate:
$dm−3K_a=1.74 \times 10^{-5}\ \text{mol dm}^{-3}$

Exam technique

In the exam

Decide what the question gives you: $K_a$ and concentration usually means find pH; pH and concentration usually means find $K_a$ ; $pK_a$ means use logs.
For weak acid pH calculations, start from $Ka=[H+][A−][HA]K_a=\frac{[\text{H}^+][\text{A}^-]}{[\text{HA}]}$ and use $[H+]=[A−][\text{H}^+]=[\text{A}^-]$ for a monoprotic acid.
Give units carefully: $K_a$ is usually mol dm⁻³ for HA calculations, while pH and $pK_a$ have no units.

Self review

Check yourself

Why does 0.100 mol dm⁻³ ethanoic acid have a higher pH than 0.100 mol dm⁻³ hydrochloric acid?
A weak acid has concentration 0.200 mol dm⁻³ and $dm−3K_a=4.0 \times 10^{-6}\ \text{mol dm}^{-3}$ . How would you estimate its pH?
Why is pH equal to $pK_a$ at half-neutralisation?

What you'll learn

What “weak” means for acids and bases in aqueous solution.
How to write a $K_a$ expression for a weak acid.
How to calculate pH from $K_a$ and concentration, and calculate $K_a$ from pH.
How to convert between $K_a$ and $pK_a$ , including the half-neutralisation method.

The prerequisites: acids, bases and pH

A Brønsted–Lowry acid is a proton donor. A Brønsted–Lowry base is a proton acceptor. A proton is a hydrogen ion, written as $H+\text{H}^+$ .

In aqueous solution, $H+\text{H}^+$ does not really float around alone — it bonds to water to form $H3O+\text{H}_3\text{O}^+$ . At A-Level, we usually write $H+\text{H}^+$ as a shorthand.

Definition

pH is a logarithmic measure of hydrogen ion concentration:

pH=−log⁡10[H+]\text{pH}=-\log_{10}[\text{H}^+]

In calculations, use the numerical value of $[H+][\text{H}^+]$ in mol dm⁻³.

Because pH is logarithmic, a change of 1 pH unit means a factor of 10 change in $[H+][\text{H}^+]$ .

Example

Converting pH into hydrogen ion concentration

A solution has pH 3.25. Find $[H+][\text{H}^+]$ .

Rearrange the pH equation by reversing the log:
$[H+]=10−pH[\text{H}^+]=10^{-\text{pH}}$
Substitute the pH value:
$[H+]=10−3.25[\text{H}^+]=10^{-3.25}$
Evaluate and include units:
$dm−3[\text{H}^+]=5.62 \times 10^{-4}\ \text{mol dm}^{-3}$

Strong, weak, concentrated and dilute

Dissociation means a substance separates into ions in solution. An aqueous solution is a solution in water.

A strong acid dissociates completely in water. For example, hydrochloric acid effectively produces one mole of $H+\text{H}^+$ for every mole of HCl dissolved.

A weak acid dissociates only slightly in water. Most acid particles remain as undissociated molecules.

Definition

Weak acid

A weak acid is an acid that only partially dissociates in aqueous solution, setting up a reversible equilibrium between the acid and its ions.

A weak base also forms ions only slightly in aqueous solution. For example, ammonia reacts only partially with water:

NH3(aq)+H2O(l)⇌NH4+(aq)+OH−(aq)\text{NH}_3\text{(aq)}+\text{H}_2\text{O(l)} \rightleftharpoons \text{NH}_4^+\text{(aq)}+\text{OH}^-\text{(aq)}

This section focuses mainly on $K_a$ , the equilibrium constant for weak acids.

Common Mistake

Weak does not mean dilute

Weak describes the extent of dissociation. Dilute describes low concentration. You can have a concentrated weak acid, or a dilute strong acid.

The weak acid equilibrium

Most A-Level weak acid calculations use a general weak monoprotic acid, HA. Monoprotic means each acid molecule can donate one proton.

The equilibrium is:

HA(aq)⇌H+(aq)+A−(aq)\text{HA(aq)} \rightleftharpoons \text{H}^+\text{(aq)}+\text{A}^-\text{(aq)}

The ion $A−\text{A}^-$ is the conjugate base of HA. A conjugate base is what remains after an acid has donated a proton.

An equilibrium is a dynamic situation where the forward and reverse reactions continue, but the concentrations of substances stay constant.

Key Idea

Weak acids are equilibrium systems

For a weak acid, HA, both undissociated acid molecules and ions are present at equilibrium. You must use equilibrium concentrations in $K_a$ expressions.

The acid dissociation constant, $K_a$

The acid dissociation constant, $K_a$ , measures how far a weak acid dissociates in water.

For:

HA(aq)⇌H+(aq)+A−(aq)\text{HA(aq)} \rightleftharpoons \text{H}^+\text{(aq)}+\text{A}^-\text{(aq)}

the expression is:

Ka=[H+][A−][HA]K_a=\frac{[\text{H}^+][\text{A}^-]}{[\text{HA}]}

Square brackets mean concentration in mol dm⁻³ at equilibrium.

For this type of expression, the units of $K_a$ are:

dm−3\frac{(\text{mol dm}^{-3})(\text{mol dm}^{-3})}{\text{mol dm}^{-3}}=\text{mol dm}^{-3}

Example

Constructing a Ka expression

Write the $K_a$ expression for ethanoic acid.

Write the dissociation equilibrium:
$CH3COOH(aq)⇌H+(aq)+CH3COO−(aq)\text{CH}_3\text{COOH(aq)} \rightleftharpoons \text{H}^+\text{(aq)}+\text{CH}_3\text{COO}^-\text{(aq)}$
Put the product concentrations on the top and the undissociated acid concentration on the bottom:
$Ka=[H+][CH3COO−][CH3COOH]K_a=\frac{[\text{H}^+][\text{CH}_3\text{COO}^-]}{[\text{CH}_3\text{COOH}]}$
Work out the units from the concentration terms:
$dm−3K_a=\text{mol dm}^{-3}$

Calculating the pH of a weak acid

Suppose the initial concentration of a weak acid HA is $c$ mol dm⁻³.

At equilibrium, let the concentration that dissociates be $x$ mol dm⁻³.

So:

$[H+]=x[\text{H}^+]=x$
$[A−]=x[\text{A}^-]=x$
$[HA]=c−x[\text{HA}]=c-x$

The exact expression is:

Ka=x2c−xK_a=\frac{x^2}{c-x}

For a weak acid, only a small amount dissociates, so $x$ is usually much smaller than $c$ . This lets you use the approximation:

\approx c

So:

Ka≈x2cK_a \approx \frac{x^2}{c}

and therefore:

[H+]≈Kac[\text{H}^+] \approx \sqrt{K_a c}

Key Idea

The square-root shortcut

For a weak monoprotic acid of concentration $c$ , if dissociation is small:

[H+]≈Kac[\text{H}^+] \approx \sqrt{K_a c}

Then use $pH=−log⁡10[H+]\text{pH}=-\log_{10}[\text{H}^+]$ .

Example

Calculating pH from Ka and concentration

Ethanoic acid has $dm−3K_a=1.74 \times 10^{-5}\ \text{mol dm}^{-3}$ . Calculate the pH of 0.100 mol dm⁻³ ethanoic acid.

Use the weak acid approximation:
$[H+]≈Kac[\text{H}^+] \approx \sqrt{K_a c}$
Substitute the values:
$[H+]≈(1.74×10−5)(0.100)[\text{H}^+] \approx \sqrt{(1.74 \times 10^{-5})(0.100)}$
Calculate the hydrogen ion concentration:
$dm−3[\text{H}^+] \approx \sqrt{1.74 \times 10^{-6}}=1.32 \times 10^{-3}\ \text{mol dm}^{-3}$
Convert to pH:
$pH=−log⁡10(1.32×10−3)=2.88\text{pH}=-\log_{10}(1.32 \times 10^{-3})=2.88$
Check the approximation is sensible:
$1.32×10−30.100×100=1.32%\frac{1.32 \times 10^{-3}}{0.100} \times 100=1.32\%$
Only a small percentage dissociated, so the approximation is reasonable.

Common Mistake

Using the starting concentration as [H+]

For a weak acid, the initial acid concentration $c$ is not the hydrogen ion concentration. Only a small fraction dissociates, so $[H+][\text{H}^+]$ must be found using $K_a$ .

Common Mistake

When the shortcut may not work

If the acid is not very weak, or the solution is extremely dilute, $\approx c$ may be a poor approximation. Then use the exact expression $Ka=x2c−xK_a=\frac{x^2}{c-x}$ instead.

Calculating $K_a$ from pH

If you know the pH and the initial concentration of a weak acid, you can work backwards.

First convert pH into $[H+][\text{H}^+]$ . For a weak monoprotic acid:

[A−]=[H+][\text{A}^-]=[\text{H}^+]

Then substitute into:

Ka=[H+][A−][HA]K_a=\frac{[\text{H}^+][\text{A}^-]}{[\text{HA}]}

Example

Calculating Ka from pH

A 0.0500 mol dm⁻³ weak monoprotic acid has pH 3.10. Calculate $K_a$ .

Convert pH to hydrogen ion concentration:
$dm−3[\text{H}^+]=10^{-3.10}=7.94 \times 10^{-4}\ \text{mol dm}^{-3}$
Use the weak acid relationships:
$dm−3[\text{A}^-]=7.94 \times 10^{-4}\ \text{mol dm}^{-3}$
and, using the small dissociation approximation:
$dm−3[\text{HA}] \approx 0.0500\ \text{mol dm}^{-3}$
Substitute into the $K_a$ expression:
$Ka=(7.94×10−4)(7.94×10−4)0.0500K_a=\frac{(7.94 \times 10^{-4})(7.94 \times 10^{-4})}{0.0500}$
Calculate:
$dm−3K_a=1.26 \times 10^{-5}\ \text{mol dm}^{-3}$

$pK_a$

$pK_a$ is another way of expressing acid strength. It is especially useful because many $K_a$ values are very small numbers written in standard form.

Definition

pKa

$pK_a$ is defined as:

pK_a=-\log_{10}K_a

It has no units.

To convert back:

K_a=10^{-pK_a}

A larger $K_a$ means a stronger weak acid. A smaller $pK_a$ also means a stronger weak acid.

Tip

Comparing acid strength

Think: large $K_a$ , low $pK_a$ , strong acid.

Example

Converting between Ka and pKa

Convert between $K_a$ and $pK_a$ .

For $dm−3K_a=6.3 \times 10^{-5}\ \text{mol dm}^{-3}$ :
$pKa=−log⁡10(6.3×10−5)=4.20pK_a=-\log_{10}(6.3 \times 10^{-5})=4.20$
For $pK_a=3.50$ :
$dm−3K_a=10^{-3.50}=3.16 \times 10^{-4}\ \text{mol dm}^{-3}$
Compare the two acids: the acid with $pK_a=3.50$ is stronger because it has the larger $K_a$ .

Finding $K_a$ from half-neutralisation

The half-neutralisation point is the point in a titration where exactly half the original acid has been neutralised by the added base.

For a weak acid HA titrated with a strong alkali such as NaOH:

HA+OH−→A−+H2O\text{HA}+\text{OH}^- \to \text{A}^-+\text{H}_2\text{O}

At half-neutralisation, half of the original HA has been converted into $A−\text{A}^-$ . Therefore:

[HA]=[A−][\text{HA}]=[\text{A}^-]

Substitute this into the $K_a$ expression:

Ka=[H+][A−][HA]K_a=\frac{[\text{H}^+][\text{A}^-]}{[\text{HA}]}

Since $[A−]=[HA][\text{A}^-]=[\text{HA}]$ , they cancel:

Ka=[H+]K_a=[\text{H}^+]

So:

pH=pKa\text{pH}=pK_a

This is why measuring the pH at half-neutralisation is a practical method for finding $K_a$ of a weak acid.

Weak acid strong base titration curve showing half-neutralisation where pH equals pKa

Example

Using half-neutralisation to find Ka

A weak monoprotic acid is titrated with NaOH. The equivalence point occurs after 30.0 cm³ of NaOH is added. At 15.0 cm³, the measured pH is 4.76. Calculate $K_a$ .

Identify the half-neutralisation volume:
$cm3\frac{30.0}{2}=15.0\ \text{cm}^3$
At half-neutralisation, use the relationship:
$pH=pKa\text{pH}=pK_a$
so:
$pK_a=4.76$
Convert $pK_a$ into $K_a$ :
$K_a=10^{-4.76}$
Calculate:
$dm−3K_a=1.74 \times 10^{-5}\ \text{mol dm}^{-3}$

Exam technique

In the exam

Decide what the question gives you: $K_a$ and concentration usually means find pH; pH and concentration usually means find $K_a$ ; $pK_a$ means use logs.
For weak acid pH calculations, start from $Ka=[H+][A−][HA]K_a=\frac{[\text{H}^+][\text{A}^-]}{[\text{HA}]}$ and use $[H+]=[A−][\text{H}^+]=[\text{A}^-]$ for a monoprotic acid.
Give units carefully: $K_a$ is usually mol dm⁻³ for HA calculations, while pH and $pK_a$ have no units.

Self review

Check yourself

Why does 0.100 mol dm⁻³ ethanoic acid have a higher pH than 0.100 mol dm⁻³ hydrochloric acid?
A weak acid has concentration 0.200 mol dm⁻³ and $dm−3K_a=4.0 \times 10^{-6}\ \text{mol dm}^{-3}$ . How would you estimate its pH?
Why is pH equal to $pK_a$ at half-neutralisation?

What you'll learn

The prerequisites: acids, bases and pH

Strong, weak, concentrated and dilute

The weak acid equilibrium

The acid dissociation constant, $K_a$

Calculating the pH of a weak acid

Calculating $K_a$ from pH

$pK_a$

Finding $K_a$ from half-neutralisation

Review Weak acids and bases $K_a$ for weak acids (A-level only) by teaching Genie

Recap your knowledge with an interactive lesson

Remember key concepts with flashcards

Weak acids and bases $K_a$ for weak acids (A-level only)

What you'll learn

The prerequisites: acids, bases and pH

Strong, weak, concentrated and dilute

The weak acid equilibrium

The acid dissociation constant, $K_a$

Calculating the pH of a weak acid

Calculating $K_a$ from pH

$pK_a$

Finding $K_a$ from half-neutralisation

Review Weak acids and bases $K_a$ for weak acids (A-level only) by teaching Genie

Recap your knowledge with an interactive lesson

Remember key concepts with flashcards

Weak acids and bases $K_a$ for weak acids (A-level only)

What you'll learn

The prerequisites: acids, bases and pH

Strong, weak, concentrated and dilute

The weak acid equilibrium

The acid dissociation constant, KaK_aKa​

Calculating the pH of a weak acid

Calculating KaK_aKa​ from pH

pKapK_apKa​

Finding KaK_aKa​ from half-neutralisation

Review Weak acids and bases $K_a$ for weak acids (A-level only) by teaching Genie

Recap your knowledge with an interactive lesson

Remember key concepts with flashcards

Weak acids and bases $K_a$ for weak acids (A-level only) Revision Guide

Weak acids and bases $K_a$ for weak acids (A-level only)

What you'll learn

The prerequisites: acids, bases and pH

Strong, weak, concentrated and dilute

The weak acid equilibrium

The acid dissociation constant, KaK_aKa​

Calculating the pH of a weak acid

Calculating KaK_aKa​ from pH

pKapK_apKa​

Finding KaK_aKa​ from half-neutralisation

Review Weak acids and bases $K_a$ for weak acids (A-level only) by teaching Genie

Recap your knowledge with an interactive lesson

Remember key concepts with flashcards

Weak acids and bases $K_a$ for weak acids (A-level only) Revision Guide

The acid dissociation constant, $K_a$

Calculating $K_a$ from pH

$pK_a$

Finding $K_a$ from half-neutralisation

The acid dissociation constant, $K_a$

Calculating $K_a$ from pH

$pK_a$

Finding $K_a$ from half-neutralisation