Table of contents

What is pKa?pKa tablepKa and pH — How to calculate pKa from pH?pKa and Ka — How to calculate pKa from Ka?Examples of pKa calculationFAQsThis pK_{a} calculator will help you **determine the pK _{a} value in two ways**: from a specific pH with the Henderson-Hasselbalch equation or from the acid dissociation constant (K

_{a}).

In this article, you will find information on:

- The
**definition of pK**;_{a} - How to use the
**pK**;_{a}table - Relationship between
**pK**;_{a}and pH - Relationship between
**pK**; and_{a}and K_{a} **Useful examples**are also provided to help you calculate pK_{a}values like a pro!

## What is pKa?

You have probably seen the term pKa before from your chemistry class in high school🧑🔬. What is it exactly? Do you recall?

It's very simple! **The pK _{a} determines how weak or strong an acid is.** To be more precise, pK

_{a}tells you how strongly the Brønsted acid holds on a given proton(H

^{+}). It allows you to predict how each acid and base solution will react in a specific experimental setting.

#### Interpretation of pK_{a}

**The lower the pK _{a}, the stronger the acid.** This means:

- The H
^{+}is held more loosely by the acid; and - The acid can give up on H
^{+}more easily.

**The higher the pK _{a}, the weaker the acid.** This means:

- The H
^{+}is held more tightly by the acid; and - The acid does not easily donate a H
^{+}.

💡 If you are looking for a way to calculate the pH of a solution in your experiment ⚗️, knowing the concentration of the acid or base is very important. You might find Omni's pH calculator and concentration calculator helpful!

## pKa table

Before going through equations and calculations of pK_{a}, here is **the easiest way to find the pK _{a} of a compound — by using the pK_{a} table.**

This table can be used when you are trying to make a buffer or carry out a reaction from scratch. It **guides the selection of acids and bases in a reaction**, since knowing the pK_{a} values of each compound will predict their reactive behavior 🧪. So, if you don't know the pH or the K_{a} of your compounds, check it out!

You can find a shortened version of the **pK _{a} table with the most common functional groups** used in basic chemistry below:

Functional group | Formula | pK |
---|---|---|

Hydroiodic acid | HI | -10 |

Hydrobromic acid | HBr | -9 |

Hydrochloric acid | HCl | -6 |

Sulfuric acid | H | -3 |

Hydronium ion | H | -1.7 |

Sulfonic acids | R–SO | -1 |

Hydrofluoric acid | HF | 3.2 |

Carboxylic acid | R–COOH | 4 |

Protonated amines | R–NH | 9-11 |

Thiols | R–SH | 13 |

Malonates | CH | 13 |

Water | H | 14 |

Alcohol | CH | 17 |

Ketone / Aldehyde | R–CH | 20-24 |

Nitrile | R–C≡N | 25 |

Ester | R–COO–R' | 25 |

Alkyne | R–C≡C–R' | 25 |

Amines | R–NH | ~35 |

Hydrogen | H | 36 |

Alkene | R–C=C–R' | ~43 |

Alkane | C | ~50 |

For more visualization, check out Master Organic Chemistry website, which provides a clear structural illustration of each functional group and the conjugate base.

## pKa and pH — How to calculate pKa from pH?

#### Henderson-Hasselbalch equation

The pK_{a} calculator is based on the well-known Henderson-Hasselbalch equation, providing the relationship between pH and pK_{a}.

$\text{pH} = \mathrm{pK_a} + \mathrm{log_{10}\frac{[A^-]}{[HA]}}$pH=pKa+log10[HA][A−]

where:

- $\mathrm{A^-}$A− — Molar concentration of the conjugate base; and
- $\mathrm{HA}$HA — Molar concentration of the weak acid.

#### Relationship of pK_{a} and pH

If $\small\mathrm{[HA] = [A^-]}$[HA]=[A−], then $\mathrm{log\large\frac{[A^-]}{[HA]} \small = 0}$log[HA][A−]=0:

When molar concentrations of weak acid and conjugate base are the same, the logarithm is exactly 0. This means that $\small\mathrm{pH = pK_a}$pH=pKa.

If $\small\mathrm{[HA] > [A^-]}$[HA]>[A−], then $\mathrm{log\large\frac{[A^-]}{[HA]}\small < 0}$log[HA][A−]<0:

When the molar concentration of the weak acid is higher than that of the conjugate base, the logarithm is negative. Thus, $\small\mathrm{pH < pK_a}$pH<pKa.

If $\small\mathrm{[HA] < [A^-]}$[HA]<[A−], then $\mathrm{log\large\frac{[A^-]}{[HA]}\small > 0}$log[HA][A−]>0:

When the molar concentration of the conjugate base is higher than that of the weak acid, the logarithm is positive. Thus, $\small\mathrm{pH > pK_a}$pH>pKa.

💡 Did you know that the **buffer capacity increases as the value of pH and pK _{a} are closer together**? This allows a buffer to maintain its pH range despite the addition of a stronger acid or base. Check our buffer capacity calculator for more information! 💬

## pKa and Ka — How to calculate pKa from Ka?

Here is another terminology to recall from your chemistry lecture🤓 — the **acid dissociation constant (K _{a})**, also known as

**acid ionization constant**.

K_{a} is a constant value measured at equilibrium, indicating how acids dissociate in a solution. **The higher K _{a} values, the stronger the acid** and the easier the dissociation (H

^{+}donating) from the other components.

#### Relationship of pK_{a} and K_{a}

pK_{a} is negatively correlated to K_{a}, meaning that if one value increases⬆️, the other value decreases⬇️. Basically, **K _{a} is simply the logarithm of pK_{a}**:

$\rm {pK_a} = {-log_{10}}[{K_a}]$pKa=−log10[Ka]

You can also write the equation as follows in case you want to **calculate an unknown K _{a} from pK_{a}**:

$\rm {K_a} = 10^{-pK_a}$Ka=10−pKa

If you are interested in knowing more about what is a logarithm and how to solve the log of a value, the log calculator will be very helpful!

💡 Unlike pH, K_{a} does not vary with the concentration, but it does vary with temperature changes. That means **K _{a} values of acids are usually fixed**.

#### Acid dissociation equation

This equation explains the **dissociation of an acid at equilibrium, giving hydronium ions (H ^{+}) and conjugate base (A^{-}) as products**. Knowing how to write the reaction in this form is very important to make sure you know which compound is the nominator or denominator when using the Henderson-Hasselbalch equation to calculate pK

_{a}or pH.

$\rm HA \rightleftharpoons H^{+} + A^{-}$HA⇌H++A−

Hence, K_{a} can be represented by the concentration of products to the concentration of the reactant:

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

## Examples of pKa calculation

✏️**Practice makes progress!** Explore the provided examples below to improve your knowledge of pK_{a} even more.

#### Calculating pK_{a} from pH

**Example 1** — Calculate the pK_{a} of a solution containing $\small0.1\ \mathrm{M}$0.1M of acetic acid and $\small0.01\ \mathrm{M}$0.01M of acetate ion. Note that $\small\mathrm{pH} = 4.8$pH=4.8.

**1. Write the acid dissociation equation** of this reaction to identify the weak acid and conjugate base:

$\footnotesize\begin{align*}\rm HA\ &\rm\rightleftharpoons H^{+} + A^{-} \\\rm CH_3COOH\ &\rm\rightleftharpoons H^{+}_3 + CH_3COO^{-} \end{align*}$HACH3COOH⇌H++A−⇌H3++CH3COO−

**2. Calculate the pK _{a}** using the Henderson-Hasselbalch equation:

$\footnotesize\begin{align*}\rm \text{pH}\ &\rm= \mathrm{pK_a} + \mathrm{log_{10}\frac{[CH_3COO^-]}{[CH_3COOH]}} \\[1.5em]-\rm{pK_a}\ &\rm= \mathrm{log_{10}\frac{[CH_3COO^-]}{[CH_3COOH]}} - \mathrm{pH} \\[1.5em]-\rm{pK_a}\ &\rm= \mathrm{log_{10}\frac{[0.01]}{[0.1]}} - 4.8 \\[1.5em]-\rm{pK_a}\ &\rm= -1 - 4.8 \\[.7em]-\rm{pK_a}\ &\rm= -5.8 \\[.7em]\rm{pK_a}\ &\rm= 5.8\end{align*}$pH−pKa−pKa−pKa−pKapKa=pKa+log10[CH3COOH][CH3COO−]=log10[CH3COOH][CH3COO−]−pH=log10[0.1][0.01]−4.8=−1−4.8=−5.8=5.8

**Example 2** — Calculate the pK_{a} of a solution containing $\small0.75\ \mathrm{M}$0.75M of lactic acid and $\small0.25\ \mathrm{M}$0.25M of sodium lactate. Note that $\small\mathrm{pH} = 3.38$pH=3.38.

**1. Write the acid dissociation equation** of this reaction to identify the weak acid and conjugate base:

$\footnotesize\begin{align*}\rm HA\ &\rm\rightleftharpoons H^{+} + A^{-} \\\rm HC_3H_5O_3\ &\rm\rightleftharpoons H^{+} + C_3H_5O^{-}_3\end{align*}$HAHC3H5O3⇌H++A−⇌H++C3H5O3−

**2. Calculate the pK _{a}** using the Henderson-Hasselbalch equation:

$\footnotesize\begin{align*}\text{pH}\ &\rm= \rm{pK_a} + \mathrm{log_{10}\frac{[C_3H_5O_3^-]}{[HC_3H_5O_3]}} \\[1.5em]-\rm{pK_a}\ &\rm= \rm{log_{10}\frac{[C_3H_5O_3^-]}{[HC_3H_5O_3]}} - \rm{pH} \\[1.5em]-\rm{pK_a}\ &\rm= \rm{log_{10}\frac{[0.25]}{[0.75]}} - 3.38 \\[1.5em]-\rm{pK_a}\ &\rm= -0.047 - 3.38 \\[.7em]-\rm{pK_a}\ &\rm= -3.85 \\[.7em]\rm{pK_a}\ &\rm= 3.85\end{align*}$pH−pKa−pKa−pKa−pKapKa=pKa+log10[HC3H5O3][C3H5O3−]=log10[HC3H5O3][C3H5O3−]−pH=log10[0.75][0.25]−3.38=−0.047−3.38=−3.85=3.85

Phew! 😮💨 That was a bit tiring, wasn't it? Let Omni Calculator's pK_{a} calculator make things a lot easier for you — easily fill in your given pH, conjugate base concentration and weak acid concentration in the section **pK _{a}frompH**.

#### Calculating pK_{a} from K_{a}

**Example 3** — If $\small\mathrm{K_a} = 1.5 × 10^{-5}$Ka=1.5×10−5, how much is pK_{a}?

$\footnotesize\begin{align*}\mathrm{pK_a} &= \mathrm{-log_{10}}[\mathrm{K_a}] \\&= \mathrm{-log_{10}}[\mathrm1.5 × 10^{-5}] \\&= 3.824\end{align*}$pKa=−log10[Ka]=−log10[1.5×10−5]=3.824

**Example 4** — If $\small\mathrm{K_a} = 6.8 × 10^{-10}$Ka=6.8×10−10, how much is pK_{a}?

$\footnotesize\begin{align*}\mathrm{pK_a} &= \mathrm{-log_{10}}[\mathrm{K_a}] \\&= \mathrm{-log_{10}}[\mathrm 6.8 × 10^{-10}] \\&= 9.162\end{align*}$pKa=−log10[Ka]=−log10[6.8×10−10]=9.162

Otherwise, feel free to input the K_{a} value in the section **pK _{a}fromK_{a}** in our pK

_{a}calculator and get your result in one second!

### What is the difference between pKa and Ka?

**K _{a} is the acid dissociation constant**, which determines how strong an acid is by its ability to dissociate in a solution.

**pK**, on the other hand, is basically the

_{a}**negative log of K**. Both of these values can determine how strong or weak an acid is.

_{a}### Are pH and pKa the same?

**No, pH and pK _{a} are two different things.** pH is a scale that measures the presence of H

^{+}ions in a solution, making it acidic, neutral, or basic. As for pK

_{a}, it tells us how strong an acid is.

### How do I calculate pKa from pH?

**Use the Henderson-Hasselbalch equation** to calculate pK_{a} from pH:

**pH = pK _{a} + log_{10}[A^{-}]/[HA]**

where **[A ^{-}]** is the conjugate base and

**[HA]**is the weak acid. The pH and molar concentrations of the acid and base must be known to calculate pK

_{a}.

### How do I calculate pKa from Ka?

Use the relationship between pK_{a} and the acid dissociation constant (K_{a}): **pK _{a} = -log_{10}[K_{a}]**. The equation can also be reverted in case pK

_{a}is given to calculate K

_{a}:

**K**.

_{a}= 10^{-pKa}### How do I calculate pKa of 0.58 M sodium acetate and 1.0 M acetic acid?

To calculate pK_{a} from pH:

Apply the Henderson-Hasselbalch equation:

**pH = pK**_{a}+ log_{10}[A^{-}]/[HA]The conjugate base, sodium acetate, is [A

^{-}]: [C_{2}H_{3}NaO_{2}]The weak acid, acetic acid, is [HA]: [CH

_{3}COOH].Thus, considering pH = 4.5, we can calculate the pK

_{a}as follows:pH = pK

_{a}+ log[C_{2}H_{3}NaO_{2}] / [CH_{3}COOH]4.5 = pK

_{a}+ log[0.58]/[1.0]pK

_{a}= -0.236 − 4.5 = 4.737Congratulations! Now, you can check your answer using Omni Calculator's pK

_{a}calculator.

### How much is pKa of acetic acid if the Ka is 1.8×10⁻⁵?

**The pK _{a} of acetic acid is 4.745**. Given the K

_{a}of 1.8 × 10

^{-5}, pK

_{a}can be calculated as follows:

pK_{a} = -log_{10}[K_{a}]

pK_{a} = -(-4.745) = 4.745

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