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 open bracket, start text, H, end text, start superscript, plus, end superscript, close bracket and start text, p, H, end text using the following equations:
- We can convert between open bracket, start text, O, H, end text, start superscript, minus, end superscript, close bracket and start text, p, O, H, end text using the following equations:
- For any aqueous solution at 25, degrees, start text, C, end text:
start text, p, H, end text, plus, start text, p, O, H, end text, equals, 14.
- For every factor of 10 increase in concentration of open bracket, start text, H, end text, start superscript, plus, end superscript, close bracket, start text, p, H, end text will decrease by 1 unit, and vice versa.
- Both acid strength and concentration determine open bracket, start text, H, end text, start superscript, plus, end superscript, close bracket and start text, p, H, end text.
Introduction
In aqueous solution, an acid is defined as any species that increases the concentration of start text, H, end text, start superscript, plus, end superscript, left parenthesis, a, q, right parenthesis, while a base increases the concentration of start text, O, H, end text, start superscript, minus, end superscript, left parenthesis, a, q, right parenthesis. 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, degrees, start text, C, end text contains 1, point, 0, times, 10, start superscript, minus, 7, end superscript, start text, space, M, end text of start text, H, end text, start superscript, plus, end superscript and start text, O, H, end text, start superscript, minus, end superscript. In comparison, the concentration of start text, H, end text, start superscript, plus, end superscript in stomach acid can be up to approximately 1, point, 0, times, 10, start superscript, minus, 1, end superscript, start text, M, end text. That means open bracket, start text, H, end text, start superscript, plus, end superscript, close bracket 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 start text, p, H, end text or start text, p, O, H, end text values. Let's look at the definitions of start text, p, H, end text and start text, p, O, H, end text.
Definitions of start text, p, H, end text and start text, p, O, H, end text
Relating open bracket, start text, H, end text, start superscript, plus, end superscript, close bracket and start text, p, H, end text
The start text, p, H, end text for an aqueous solution is calculated from open bracket, start text, H, end text, start superscript, plus, end superscript, close bracket using the following equation:
The lowercase start text, p, end text indicates `, `, minus, start text, l, o, g, end text, start subscript, 10, end subscript, ". You will often see people leave off the base 10 part as an abbreviation.
For example, if we have a solution with open bracket, start text, H, end text, start superscript, plus, end superscript, close bracket, equals, 1, times, 10, start superscript, minus, 5, end superscript, start text, space, M, end text, then we can calculate the start text, p, H, end text using Eq. 1a:
Given the start text, p, H, end text of a solution, we can also find open bracket, start text, H, end text, start superscript, plus, end superscript, close bracket:
Relating open bracket, start text, O, H, end text, start superscript, minus, end superscript, close bracket and start text, p, O, H, end text
The start text, p, O, H, end text for an aqueous solution is defined in the same way for open bracket, start text, O, H, end text, start superscript, minus, end superscript, close bracket:
For example, if we have a solution with open bracket, start text, O, H, end text, start superscript, minus, end superscript, close bracket, equals, 1, times, 10, start superscript, minus, 12, end superscript, start text, space, M, end text, then we can calculate start text, p, O, H, end text using Eq. 2a:
Given the start text, p, O, H, end text of a solution, we can also find open bracket, start text, O, H, end text, start superscript, minus, end superscript, close bracket:
Relating start text, p, H, end text and start text, p, O, H, end text
Based on equilibrium concentrations of start text, H, end text, start superscript, plus, end superscript and start text, O, H, end text, start superscript, minus, end superscript in water, the following relationship is true for any aqueous solution at 25, degrees, start text, C, end text:
This relationship can be used to convert between start text, p, H, end text and start text, p, O, H, end text. In combination with Eq. 1a/b and Eq. 2a/b, we can always relate start text, p, O, H, end text and/or start text, p, H, end text to open bracket, start text, O, H, end text, start superscript, minus, end superscript, close bracket and open bracket, start text, H, end text, start superscript, plus, end superscript, close bracket. For a derivation of this equation, check out the article on the autoionization of water.
Example 1: Calculating the start text, p, H, end text of a strong base solution
If we use 1, point, 0, start text, space, m, m, o, l, end text of start text, N, a, O, H, end text to make 1, point, 0, start text, space, L, end text of an aqueous solution at 25, degrees, start text, C, end text, what is the start text, p, H, end text of this solution?
We can find the start text, p, H, end text of our start text, N, a, O, H, end text solution by using the relationship between open bracket, start text, O, H, end text, start superscript, minus, end superscript, close bracket, start text, p, H, end text, and start text, p, O, H, end text. Let's go through the calculation step-by-step.
Step 1. Calculate the molar concentration of start text, N, a, O, H, end text
Molar concentration is equal to moles of solute per liter of solution:
To calculate the molar concentration of start text, N, a, O, H, end text, we can use the known values for the moles of start text, N, a, O, H, end text and the volume of solution:
The concentration of start text, N, a, O, H, end text in the solution is 1, point, 0, times, 10, start superscript, minus, 3, end superscript, start text, space, M, end text.
Step 2: Calculate open bracket, start text, O, H, end text, start superscript, minus, end superscript, close bracket based on the dissociation of start text, N, a, O, H, end text
Because start text, N, a, O, H, end text is a strong base, it dissociates completely into its constituent ions in aqueous solution:
This balanced equation tells us that every mole of start text, N, a, O, H, end text produces one mole of start text, O, H, end text, start superscript, minus, end superscript in aqueous solution. Therefore, we have the following relationship between open bracket, start text, N, a, O, H, end text, close bracket and open bracket, start text, O, H, end text, start superscript, minus, end superscript, close bracket:
Step 3: Calculate start text, p, O, H, end text from open bracket, start text, O, H, end text, start superscript, minus, end superscript, close bracket using Eq. 2a
Now that we know the concentration of start text, O, H, end text, start superscript, minus, end superscript, we can calculate start text, p, O, H, end text using Eq. 2a:
The start text, p, O, H, end text of our solution is 3, point, 00.
Step 4: Calculate start text, p, H, end text from start text, p, O, H, end text using Eq. 3
We can calculate start text, p, H, end text from start text, p, O, H, end text using Eq. 3. Rearranging to solve for our unknown, start text, p, H, end text:
We can substitute the value of start text, p, O, H, end text we found in Step 3 to find the start text, p, H, end text:
Therefore, the start text, p, H, end text of our start text, N, a, O, H, end text solution is 11, point, 00.
The start text, p, H, end text scale: Acidic, basic, and neutral solutions
Converting open bracket, start text, H, end text, start superscript, plus, end superscript, close bracket to start text, p, H, end text is a convenient way to gauge the relative acidity or basicity of a solution. The start text, p, H, end text scale allows us to easily rank different substances by their start text, p, H, end text value.
The start text, p, H, end text scale is a negative logarithmic scale. The logarithmic part means that start text, p, H, end text changes by 1 unit for every factor of 10 change in concentration of start text, H, end text, start superscript, plus, end superscript. The negative sign in front of the log tells us that there is an inverse relationship between start text, p, H, end text and open bracket, start text, H, end text, start superscript, plus, end superscript, close bracket: when start text, p, H, end text increases, open bracket, start text, H, end text, start superscript, plus, end superscript, close bracket decreases, and vice versa.
The following image shows a start text, p, H, end text scale labeled with start text, p, H, end text values for some common household substances. These start text, p, H, end text values are for solutions at 25, degrees, start text, C, end text. Note that it is possible to have a negative start text, p, H, end text value.
Some important terminology to remember for aqueous solutions at 25, degrees, start text, C, end text:
- For a neutral solution, start text, p, H, end text, equals, 7.
- Acidic solutions have start text, p, H, end text, is less than, 7.
- Basic solutions have start text, p, H, end text, is greater than, 7.
The lower the start text, p, H, end text value, the more acidic the solution and the higher the concentration of start text, H, end text, start superscript, plus, end superscript. The higher the start text, p, H, end text value, the more basic the solution and the lower the concentration of start text, H, end text, start superscript, plus, end superscript. While we could also describe the acidity or basicity of a solution in terms of start text, p, O, H, end text, it is a little more common to use start text, p, H, end text. Luckily, we can easily convert between start text, p, H, end text and start text, p, O, H, end text values.
Concept check: Based on the start text, p, H, end text scale given above, which solution is more acidicminusorange juice, or vinegar?
Example 2: Determining the start text, p, H, end text of a diluted strong acid solution
We have 100, start text, space, m, L, end text of a nitric acid solution with a start text, p, H, end text of 4, point, 0. We dilute the solution by adding water to get a total volume of 1, point, 0, start text, space, L, end text.
What is the start text, p, H, end text 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 start text, p, H, end text scale is a negative logarithmic scale. Therefore, if the concentration of start text, H, end text, start superscript, plus, end superscript decreases by a single factor of 10, then the start text, p, H, end text will increase by 1 unit.
Since the original volume, 100, start text, space, m, L, end text, is one tenth the total volume after dilution, the concentration of start text, H, end text, start superscript, plus, end superscript in solution has been reduced by a factor of 10. Therefore, the start text, p, H, end text of the solution will increase 1 unit:
Therefore, the start text, p, H, end text of the diluted solution is 5, point, 0.
Method 2. Use moles of start text, H, end text, start superscript, plus, end superscript to calculate start text, p, H, end text
Step 1: Calculate moles of start text, H, end text, start superscript, plus, end superscript
We can use the start text, p, H, end text and volume of the original solution to calculate the moles of start text, H, end text, start superscript, plus, end superscript in the solution.
Step 2: Calculate molarity of start text, H, end text, start superscript, plus, end superscript after dilution
The molarity of the diluted solution can be calculated by using the moles of start text, H, end text, start superscript, plus, end superscript from the original solution and the total volume after dilution.
Step 3: Calculate start text, p, H, end text from open bracket, start text, H, end text, start superscript, plus, end superscript, close bracket
Finally, we can use Eq. 1a to calculate start text, p, H, end text:
Method 2 gives us the same answer as Method 1, hooray!
In general, Method 2 takes a few extra steps, but it can always be used to find changes in start text, p, H, end text. Method 1 is a handy shortcut when changes in concentration occur as multiples of 10. Method 1 can also be used as a quick way to estimate start text, p, H, end text changes.
Relationship between start text, p, H, end text and acid strength
Based on the equation for start text, p, H, end text, we know that start text, p, H, end text is related to open bracket, start text, H, end text, start superscript, plus, end superscript, close bracket. However, it is important to remember that start text, p, H, end text 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 open bracket, start text, H, end text, start superscript, plus, end superscript, close bracket at a given acid concentration. For example, a 1, point, 0, start text, M, end text solution of strong acid start text, H, C, l, end text will have a higher concentration of start text, H, end text, start superscript, plus, end superscript than a 1, point, 0, start text, M, end text solution of weak acid start text, H, F, end text. Thus, for two solutions of monoprotic acid at the same concentration, start text, p, H, end text will be proportional to acid strength.
More generally though, both acid strength and concentration determine open bracket, start text, H, end text, start superscript, plus, end superscript, close bracket. Therefore, we can't always assume that the start text, p, H, end text of a strong acid solution will be lower than the start text, p, H, end text of a weak acid solution. The acid concentration matters too!
Summary
- We can convert between open bracket, start text, H, end text, start superscript, plus, end superscript, close bracket and start text, p, H, end text using the following equations:
- We can convert between open bracket, start text, O, H, end text, start superscript, minus, end superscript, close bracket and start text, p, O, H, end text using the following equations:
- For every factor of 10 increase in concentration of open bracket, start text, H, end text, start superscript, plus, end superscript, close bracket, start text, p, H, end text will decrease by 1 unit, and vice versa.
- For any aqueous solution at 25, degrees, start text, C, end text:
start text, p, H, end text, plus, start text, p, O, H, end text, equals, 14.
- Both acid strength and concentration determine open bracket, start text, H, end text, start superscript, plus, end superscript, close bracket and start text, p, H, end text.
Problem 1: Calculating the pH of a strong base solution at 25, degrees, start text, C, end text
We make 200, start text, space, m, L, end text of a solution with a 0, point, 025, start text, space, M, end text concentration of start text, C, a, left parenthesis, O, H, right parenthesis, end text, start subscript, 2, end subscript. The solution is then diluted to 1, point, 00, start text, space, L, end text by adding additional water.
What is the start text, p, H, end text 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.(42 votes)
- What does M stand for in the unit labels?(15 votes)
- M stands for the unit of Molarity of the solution.(19 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)
- 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.(24 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.(3 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)