Main content

## Chemistry library

### Course: Chemistry library > Unit 13

Lesson 1: Acids, bases, and pH- Arrhenius acids and bases
- Arrhenius acids and bases
- pH, pOH, and the pH scale
- Brønsted-Lowry acids and bases
- Brønsted–Lowry acids and bases
- Autoionization of water
- Water autoionization and Kw
- Definition of pH
- Strong acid solutions
- Strong base solutions
- Acid strength, anion size, and bond energy
- Identifying weak acids and strong acids
- Identifying weak bases and strong bases
- Introduction to acid–base reactions

© 2023 Khan AcademyTerms of usePrivacy PolicyCookie Notice

# pH, pOH, and the pH scale

Definitions of pH, pOH, and the pH scale. Calculating the pH of a strong acid or base solution. The relationship between acid strength and the pH of a solution.

## Key points

- We can convert between
and$[{\text{H}}^{+}]$ using the following equations:$\text{pH}$

- We can convert between
and$[{\text{OH}}^{-}]$ using the following equations:$\text{pOH}$

- For any aqueous solution at
:$25{{\textstyle \phantom{\rule{0.167em}{0ex}}}}^{\circ}\text{C}$

. $\text{pH}+\text{pOH}=14$

- For every factor of
$10$ *increase*in concentration of ,$[{\text{H}}^{+}]$ will$\text{pH}$ *decrease*by unit, and vice versa.$1$ - Both acid strength and concentration determine
and$[{\text{H}}^{+}]$ .$\text{pH}$

## Introduction

In aqueous solution, an acid is defined as any species that increases the concentration of ${\text{H}}^{+}(aq)$ , while a base increases the concentration of ${\text{OH}}^{-}(aq)$ . Typical concentrations of these ions in solution can be very small, and they also span a wide range.

For example, a sample of pure water at $25{{\textstyle \phantom{\rule{0.167em}{0ex}}}}^{\circ}\text{C}$ contains $1.0\times {10}^{-7}\text{M}$ of ${\text{H}}^{+}$ and ${\text{OH}}^{-}$ . In comparison, the concentration of ${\text{H}}^{+}$ in stomach acid can be up to approximately $1.0\times {10}^{-1}{\textstyle \phantom{\rule{0.167em}{0ex}}}\text{M}$ . That means $[{\text{H}}^{+}]$ for stomach acid is approximately $6$ orders of magnitude larger than in pure water!

To avoid dealing with such hairy numbers, scientists convert these concentrations to $\text{pH}$ or $\text{pOH}$ values. Let's look at the definitions of $\text{pH}$ and $\text{pOH}$ .

## Definitions of $\text{pH}$ and $\text{pOH}$

### Relating $[{\text{H}}^{+}]$ and $\text{pH}$

The $\text{pH}$ for an aqueous solution is calculated from $[{\text{H}}^{+}]$ using the following equation:

The lowercase $\text{p}$ indicates $\u2018\u2018-{\text{log}}_{10}"$ . You will often see people leave off the base $10$ part as an abbreviation.

For example, if we have a solution with $[{\text{H}}^{+}]=1\times {10}^{-5}\text{M}$ , then we can calculate the $\text{pH}$ using

**Eq. 1a**:Given the $\text{pH}$ of a solution, we can also find $[{\text{H}}^{+}]$ :

### Relating $[{\text{OH}}^{-}]$ and $\text{pOH}$

The $\text{pOH}$ for an aqueous solution is defined in the same way for $[{\text{OH}}^{-}]$ :

For example, if we have a solution with $[{\text{OH}}^{-}]=1\times {10}^{-12}\text{M}$ , then we can calculate $\text{pOH}$ using

**Eq. 2a**:Given the $\text{pOH}$ of a solution, we can also find $[{\text{OH}}^{-}]$ :

### Relating $\text{pH}$ and $\text{pOH}$

Based on equilibrium concentrations of ${\text{H}}^{+}$ and ${\text{OH}}^{-}$ in water, the following relationship is true for any aqueous solution at $25{{\textstyle \phantom{\rule{0.167em}{0ex}}}}^{\circ}\text{C}$ :

This relationship can be used to convert between $\text{pH}$ and $\text{pOH}$ . In combination with $\text{pOH}$ and/or $\text{pH}$ to $[{\text{OH}}^{-}]$ and $[{\text{H}}^{+}]$ . For a derivation of this equation, check out the article on the autoionization of water.

**Eq. 1a/b**and**Eq. 2a/b**, we can always relate## Example 1: Calculating the $\text{pH}$ of a strong base solution

**If we use**$1.0\text{mmol}$ of $\text{NaOH}$ to make $1.0\text{L}$ of an aqueous solution at $25{{\textstyle \phantom{\rule{0.167em}{0ex}}}}^{\circ}\text{C}$ , what is the $\text{pH}$ of this solution?

We can find the $\text{pH}$ of our $\text{NaOH}$ solution by using the relationship between $[{\text{OH}}^{-}]$ , $\text{pH}$ , and $\text{pOH}$ . Let's go through the calculation step-by-step.

### Step 1. Calculate the molar concentration of $\text{NaOH}$

Molar concentration is equal to moles of solute per liter of solution:

To calculate the molar concentration of $\text{NaOH}$ , we can use the known values for the moles of $\text{NaOH}$ and the volume of solution:

The concentration of $\text{NaOH}$ in the solution is $1.0\times {10}^{-3}\text{M}$ .

### Step 2: Calculate $[{\text{OH}}^{-}]$ based on the dissociation of $\text{NaOH}$

Because $\text{NaOH}$ is a strong base, it dissociates completely into its constituent ions in aqueous solution:

This balanced equation tells us that every mole of $\text{NaOH}$ produces one mole of ${\text{OH}}^{-}$ in aqueous solution. Therefore, we have the following relationship between $[\text{NaOH}]$ and $[{\text{OH}}^{-}]$ :

### Step 3: Calculate $\text{pOH}$ from $[{\text{OH}}^{-}]$ using Eq. 2a

Now that we know the concentration of ${\text{OH}}^{-}$ , we can calculate $\text{pOH}$ using

**Eq. 2a**:The $\text{pOH}$ of our solution is $3.00$ .

### Step 4: Calculate $\text{pH}$ from $\text{pOH}$ using Eq. 3

We can calculate $\text{pH}$ from $\text{pOH}$ using $\text{pH}$ :

**Eq. 3**. Rearranging to solve for our unknown,We can substitute the value of $\text{pOH}$ we found in $\text{pH}$ :

**Step 3**to find theTherefore, the $\text{pH}$ of our $\text{NaOH}$ solution is $11.00$ .

## The $\text{pH}$ scale: Acidic, basic, and neutral solutions

Converting $[{\text{H}}^{+}]$ to $\text{pH}$ is a convenient way to gauge the relative acidity or basicity of a solution. The $\text{pH}$ scale allows us to easily rank different substances by their $\text{pH}$ value.

The $\text{pH}$ scale is a negative logarithmic scale. The logarithmic part means that $\text{pH}$ changes by $1$ unit for every factor of $10$ change in concentration of ${\text{H}}^{+}$ . The negative sign in front of the log tells us that there is an $\text{pH}$ and $[{\text{H}}^{+}]$ : when $\text{pH}$ increases, $[{\text{H}}^{+}]$ decreases, and vice versa.

*inverse relationship*betweenThe following image shows a $\text{pH}$ scale labeled with $\text{pH}$ values for some common household substances. These $\text{pH}$ values are for solutions at $25{{\textstyle \phantom{\rule{0.167em}{0ex}}}}^{\circ}\text{C}$ . Note that it is possible to have a negative $\text{pH}$ value.

Some important terminology to remember for aqueous solutions at $25{{\textstyle \phantom{\rule{0.167em}{0ex}}}}^{\circ}\text{C}$ :

- For a
*neutral*solution, .$\text{pH}=7$ *Acidic*solutions have .$\text{pH}<7$ *Basic*solutions have .$\text{pH}>7$

The lower the $\text{pH}$ value, the more acidic the solution and the higher the concentration of ${\text{H}}^{+}$ . The higher the $\text{pH}$ value, the more basic the solution and the lower the concentration of ${\text{H}}^{+}$ . While we could also describe the acidity or basicity of a solution in terms of $\text{pOH}$ , it is a little more common to use $\text{pH}$ . Luckily, we can easily convert between $\text{pH}$ and $\text{pOH}$ values.

**Concept check: Based on the**$\text{pH}$ scale given above, which solution is more acidic$-$ orange juice, or vinegar?

## Example $2$ : Determining the $\text{pH}$ of a diluted strong acid solution

We have $100\text{mL}$ of a nitric acid solution with a $\text{pH}$ of $4.0$ . We dilute the solution by adding water to get a total volume of $1.0\text{L}$ .

**What is the**$\text{pH}$ of the diluted solution?

There are multiple ways to solve this problem. We will go over two different methods.

### Method 1. Use properties of the log scale

Recall that $\text{pH}$ scale is a negative logarithmic scale. Therefore, if the concentration of ${\text{H}}^{+}$ decreases by a single factor of $10$ , then the $\text{pH}$ will $1$ unit.

*increase*bySince the original volume, $100\text{mL}$ , is one tenth the total volume after dilution, the concentration of ${\text{H}}^{+}$ in solution has been reduced by a factor of $10$ . Therefore, the $\text{pH}$ of the solution will increase $1$ unit:

Therefore, the $\text{pH}$ of the diluted solution is $5.0$ .

### Method 2. Use moles of ${\text{H}}^{+}$ to calculate $\text{pH}$

#### Step 1: Calculate moles of ${\text{H}}^{+}$

We can use the $\text{pH}$ and volume of the original solution to calculate the moles of ${\text{H}}^{+}$ in the solution.

#### Step 2: Calculate molarity of ${\text{H}}^{+}$ after dilution

The molarity of the diluted solution can be calculated by using the moles of ${\text{H}}^{+}$ from the original solution and the total volume after dilution.

#### Step 3: Calculate $\text{pH}$ from $[{\text{H}}^{+}]$

Finally, we can use $\text{pH}$ :

**Eq. 1a**to calculate**Method 2**gives us the same answer as

**Method 1**, hooray!

In general, $\text{pH}$ . $10$ . $\text{pH}$ changes.

**Method 2**takes a few extra steps, but it can always be used to find changes in**Method 1**is a handy shortcut when changes in concentration occur as multiples of**Method 1**can also be used as a quick way to estimate## Relationship between $\text{pH}$ and acid strength

Based on the equation for $\text{pH}$ , we know that $\text{pH}$ is related to $[{\text{H}}^{+}]$ . However, it is important to remember that $\text{pH}$ is

*not*always directly related to acid strength.The strength of an acid depends on the amount that the acid dissociates in solution: the stronger the acid, the higher $[{\text{H}}^{+}]$ at a given acid concentration. For example, a $1.0{\textstyle \phantom{\rule{0.167em}{0ex}}}\text{M}$ solution of strong acid $\text{HCl}$ will have a higher concentration of ${\text{H}}^{+}$ than a $1.0{\textstyle \phantom{\rule{0.167em}{0ex}}}\text{M}$ solution of weak acid $\text{HF}$ . Thus, for two solutions of monoprotic acid at the same concentration, $\text{pH}$ will be proportional to acid strength.

More generally though, both acid strength and concentration determine $[{\text{H}}^{+}]$ . Therefore, we can't always assume that the $\text{pH}$ of a strong acid solution will be lower than the $\text{pH}$ of a weak acid solution. The acid concentration matters too!

## Summary

- We can convert between
and$[{\text{H}}^{+}]$ using the following equations:$\text{pH}$

- We can convert between
and$[{\text{OH}}^{-}]$ using the following equations:$\text{pOH}$

- For every factor of
$10$ *increase*in concentration of ,$[{\text{H}}^{+}]$ will$\text{pH}$ *decrease*by unit, and vice versa.$1$ - For any aqueous solution at
:$25{{\textstyle \phantom{\rule{0.167em}{0ex}}}}^{\circ}\text{C}$

. $\text{pH}+\text{pOH}=14$

- Both acid strength and concentration determine
and$[{\text{H}}^{+}]$ .$\text{pH}$

## Problem 1: Calculating the pH of a strong base solution at $25{{\textstyle \phantom{\rule{0.167em}{0ex}}}}^{\circ}\text{C}$

We make $200\text{mL}$ of a solution with a $0.025\text{M}$ concentration of ${\text{Ca(OH)}}_{2}$ . The solution is then diluted to $1.00\text{L}$ by adding additional water.

**What is the**$\text{pH}$ of the solution after dilution?

## Want to join the conversation?

- How does the temperature affect the pH and pOH?(36 votes)
- At 100C the pH of water is 6.14, so higher temperature decreases the pH. The opposite is true for pOH: higher temperatures increases the pOH.(43 votes)

- What does M stand for in the unit labels?(15 votes)
- M stands for the unit of Molarity of the solution.(20 votes)

- Could someone explain the difference between acid strength and concentration? According to me, a strong acid will fully ionise in water compared to a weak acid which will partially ionise. Therefore a strong acid will contribute more H+ ions than a weak acid. Therefore, the pH of a strong acid solution will be higher than a weak acid solution.

Is this correct?(8 votes)- Nice question!! It is important that you don't confuse the words strong and weak with the terms concentrated and dilute. At the same concentration, a weak acid will be less acidic than a strong acid. However, if you have highly concentrated weak acid (almost pure) and compare this to a very diluted strong acid (like 1 drop of HCl in a swimming pool) then the pH of the weak acid will be much more acidic than that of the strong acid.(25 votes)

- Can someone please explain what are monoprotic and diprotic acids? Thanks.(13 votes)
- H2SO4 is a typical diprotic acid (2 protons can be released in aqueous solution, however one at that time)

H2SO4 +H20 gives HSO4- + H+/H3O+

then,

HSO4- +H2O gives SO42- + H+/H3O+(16 votes)

- how can we solve pH,pOH numericals without using scientific calculator during our examination?(8 votes)
- Unless you are using perfect numbers of base ten (e.g. 10^-7, 10^-2, etc.) there is no way to do it because you cannot easily do logarithms. If you are asked to do these calculations without a calculator, there is a good chance minimal if any extensive calculations are required.

Example: log(10^-6) = -6(7 votes)

- How can NaOH have a pH scale? How can a base add H+ ions to the solution? It adds OH- ions right?(5 votes)
- NaOH does not have a pH, but an aqueous solution of NaOH does.

Water contains both H⁺ and OH⁻ ions.

Adding NaOH increases the concentration of OH⁻ ions and decreases the concentration of H⁺ ions.

But there are**always**some H⁺ ions present, so aqueous NaOH solutions have a pH, usually between 7 and 14.(11 votes)

- How does pH+pOH= 14? Where does the random number come in?(3 votes)
- There is no random number.

The formula comes from the ion product for water.

[H⁺][OH⁻] = 1.00 × 10⁻¹⁴

∴ pH + pOH = 14.00(4 votes)

- Hi

I feel like there's a step missing.I'm not sure why the pH as an exponent is negative & where the minus sign comes from.I understand that the logarithm (of base 10) was changed to an exponent.What is this law of logarithms called?

pH= -log (H+)

10^-pH = (H+)

Also I was trying to figure it out with numbers pH=-(log 10^-4)

and I got 10^-pH =10^-4 and I'm not sure where to go from there to obtain the pH.Do I just cancel out the 10s & minuses that are on both sides to get a pH of 4,to cross off these I have to divide/multiply both sides by some number(s) would these numbers be 10 and multiplying the exponents by -1 to get rid of the minuses because the pH scale is usually positive numbers?

Thanks!(3 votes)- This is the
*power rule*of logs. When you have a number in front of a log term, this is the same as raising the log term to that number. For example, 4log(3) is the same as log(3^4).

With pH, the number in front of the log is -1 (because pH = -log [H+]). Therefore, using the power rule, we can re-express this as pH = log ([H+]^-1).

Using another log rule, we can express each side of this equation as an exponent of 10 and we get:

10^pH = 10^log ([H+]^-1).

Using a definition of logs, the right hand side of this equation now just becomes [H+]^-1. So we have:

10^pH = [H+]^-1

The right hand side can be expressed as 1/[H+] giving us:

10^pH = 1/[H+]

Multiplying each side by [H+] and dividing each side by 10^pH gives:

[H+] = 1/10^pH which is the same as saying [H+] = 10^-pH.

As an example, if the pH is 7, then [H+] = 10^-7.(3 votes)

- what is -log? is it a number?(2 votes)
- Log is a maths function, it stands for logarithm.

We use a log scale so we can visualise the concentration of H+ ions easier as we don't without have to deal with lots of decimal places etc.(4 votes)

- Is it necessary that to calculate pH or pOH of a substance, we need to have the substance in aqueous physical state? What if the substance is already in aqueous state?

Is it possible to find the pH or pOH of a substance in solid physical state or even by melting it?

Also, we always calculate the pH or pOH of a solid substance by dissolving it in water and find its pH or pOH according to the [H+] or [OH-] produced by it .Why we don't take in consideration the [H+] or [OH-] present in water and only take [H+] or [OH-]of the dissolved substance?

Thankyou :)(3 votes)- Ok so while calculating the equilibrium constant for a reaction, we consider the active masses of substances involved in the reactants and products. Since the activity of pure solids and liquids is unity, adding them to the expression will make no difference and it is usually omitted.

Hope this helps!(1 vote)