How does the temperature affect the pH and pOH?

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.

What does M stand for in the unit labels?

M stands for the unit of Molarity of the solution.

Can someone please explain what are monoprotic and diprotic acids? Thanks.

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+

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?

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.

how can we solve pH,pOH numericals without using scientific calculator during our examination?

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

How can NaOH have a pH scale? How can a base add H+ ions to the solution? It adds OH- ions right?

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.

How does pH+pOH= 14? Where does the random number come in?

There is no random number. The formula comes from the ion product for water. [H⁺][OH⁻] = 1.00 × 10⁻¹⁴ ∴ pH + pOH = 14.00

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!

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.

what is -log? is it a number?

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.

Main content

pH, pOH, and the pH scale

Google Classroom

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 $[H^{+}]$ ‍ and $pH$ ‍ using the following equations:

$\begin{aligned} pH & = - \log [H^{+}] \\ [H^{+}] & = 10^{- pH} \end{aligned}$ ‍

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

$\begin{aligned} pOH & = - \log [{OH}^{-}] \\ [{OH}^{-}] & = 10^{- pOH} \end{aligned}$ ‍

For any aqueous solution at $25^{\circ} C$ ‍:

$pH + pOH = 14$ ‍.

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

Introduction

In aqueous solution, an acid is defined as any species that increases the concentration of

H^{+} (a q)

, while a base increases the concentration of

{OH}^{-} (a q)

. 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^{\circ} C

contains

1.0 \times 10^{- 7} M

H^{+}

and

{OH}^{-}

. In comparison, the concentration of

H^{+}

in stomach acid can be up to approximately

1.0 \times 10^{- 1} M

. That means

[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

pH

pOH

values. Let's look at the definitions of

pH

and

pOH

Definitions of $pH$ ‍ and $pOH$ ‍

Relating $[H^{+}]$ ‍ and $pH$ ‍

The

pH

for an aqueous solution is calculated from

[H^{+}]

using the following equation:

$pH = - \log [H^{+}] (Eq. 1a)$ ‍

The lowercase

p

indicates

‘ ‘ - {log}_{10} "

. You will often see people leave off the base

10

part as an abbreviation.

For example, if we have a solution with

[H^{+}] = 1 \times 10^{- 5} M

, then we can calculate the

pH

using Eq. 1a:

$pH = - \log (1 \times 10^{- 5}) = 5.0$ ‍

Given the

pH

of a solution, we can also find

[H^{+}]

$[H^{+}] = 10^{- pH} (Eq. 1b)$ ‍

Relating $[{OH}^{-}]$ ‍ and $pOH$ ‍

The

pOH

for an aqueous solution is defined in the same way for

[{OH}^{-}]

$pOH = - \log [{OH}^{-}] (Eq. 2a)$ ‍

For example, if we have a solution with

[{OH}^{-}] = 1 \times 10^{- 12} M

, then we can calculate

pOH

using Eq. 2a:

$pOH = - \log (1 \times 10^{- 12}) = 12.0$ ‍

Given the

pOH

of a solution, we can also find

[{OH}^{-}]

$10^{- pOH} = [{OH}^{-}] (Eq. 2b)$ ‍

Relating $pH$ ‍ and $pOH$ ‍

Based on equilibrium concentrations of

H^{+}

and

{OH}^{-}

in water, the following relationship is true for any aqueous solution at

25^{\circ} C

pH + pOH = 14 (Eq. 3)

This relationship can be used to convert between

pH

and

pOH

. In combination with Eq. 1a/b and Eq. 2a/b, we can always relate

pOH

and/or

pH

[{OH}^{-}]

and

[H^{+}]

. For a derivation of this equation, check out the article on the autoionization of water.

Example 1: Calculating the $pH$ ‍ of a strong base solution

If we use $1.0 mmol$ ‍ of $NaOH$ ‍ to make $1.0 L$ ‍ of an aqueous solution at $25^{\circ} C$ ‍, what is the $pH$ ‍ of this solution?

We can find the

pH

of our

NaOH

solution by using the relationship between

[{OH}^{-}]

pH

, and

pOH

. Let's go through the calculation step-by-step.

Step 1. Calculate the molar concentration of $NaOH$ ‍

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

$Molar concentration = \frac{mol solute}{L solution}$ ‍

To calculate the molar concentration of

NaOH

, we can use the known values for the moles of

NaOH

and the volume of solution:

$\begin{aligned} [NaOH] & = \frac{1.0 mmol NaOH}{1.0 L} \\ = \frac{1.0 \times 10^{- 3} mol NaOH}{1.0 L} \\ = 1.0 \times 10^{- 3} M NaOH \end{aligned}$ ‍

The concentration of

NaOH

in the solution is

1.0 \times 10^{- 3} M

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

Because

NaOH

is a strong base, it dissociates completely into its constituent ions in aqueous solution:

$NaOH (a q) \to {Na}^{+} (a q) + {OH}^{-} (a q)$ ‍

This balanced equation tells us that every mole of

NaOH

produces one mole of

{OH}^{-}

in aqueous solution. Therefore, we have the following relationship between

[NaOH]

and

[{OH}^{-}]

$[NaOH] = [{OH}^{-}] = 1.0 \times 10^{- 3} M$ ‍

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

Now that we know the concentration of

{OH}^{-}

, we can calculate

pOH

using Eq. 2a:

$\begin{aligned} pOH & = - \log [{OH}^{-}] \\ = - \log (1.0 \times 10^{- 3}) \\ = 3.00 \end{aligned}$ ‍

The

pOH

of our solution is

3.00

Step 4: Calculate $pH$ ‍ from $pOH$ ‍ using Eq. 3

We can calculate

pH

from

pOH

using Eq. 3. Rearranging to solve for our unknown,

pH

pH = 14 - pOH

We can substitute the value of

pOH

we found in Step 3 to find the

pH

pH = 14 - 3.00 = 11.00

Therefore, the

pH

of our

NaOH

solution is

11.00

The $pH$ ‍ scale: Acidic, basic, and neutral solutions

Converting

[H^{+}]

pH

is a convenient way to gauge the relative acidity or basicity of a solution. The

pH

scale allows us to easily rank different substances by their

pH

value.

The

pH

scale is a negative logarithmic scale. The logarithmic part means that

pH

changes by

1

unit for every factor of

10

change in concentration of

H^{+}

. The negative sign in front of the log tells us that there is an inverse relationship between

pH

and

[H^{+}]

: when

pH

increases,

[H^{+}]

decreases, and vice versa.

The following image shows a

pH

scale labeled with

pH

values for some common household substances. These

pH

values are for solutions at

25^{\circ} C

. Note that it is possible to have a negative

pH

value.

Some important terminology to remember for aqueous solutions at

25^{\circ} C

For a neutral solution, $pH = 7$ ‍.
Acidic solutions have $pH < 7$ ‍.
Basic solutions have $pH > 7$ ‍.

The lower the

pH

value, the more acidic the solution and the higher the concentration of

H^{+}

. The higher the

pH

value, the more basic the solution and the lower the concentration of

H^{+}

. While we could also describe the acidity or basicity of a solution in terms of

pOH

, it is a little more common to use

pH

. Luckily, we can easily convert between

pH

and

pOH

values.

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

Example $2$ ‍: Determining the $pH$ ‍ of a diluted strong acid solution

We have

100 mL

of a nitric acid solution with a

pH

4.0

. We dilute the solution by adding water to get a total volume of

1.0 L

What is the $pH$ ‍ 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

pH

scale is a negative logarithmic scale. Therefore, if the concentration of

H^{+}

decreases by a single factor of

10

, then the

pH

will increase by

1

unit.

Since the original volume,

100 mL

, is one tenth the total volume after dilution, the concentration of

H^{+}

in solution has been reduced by a factor of

10

. Therefore, the

pH

of the solution will increase

1

unit:

$\begin{aligned} pH & = original pH + 1.0 \\ = 4.0 + 1.0 \\ = 5.0 \end{aligned}$ ‍

Therefore, the

pH

of the diluted solution is

5.0

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

Step 1: Calculate moles of $H^{+}$ ‍

We can use the

pH

and volume of the original solution to calculate the moles of

H^{+}

in the solution.

$\begin{aligned} {moles H}^{+} & = [H^{+}]_{i n i t i a l} \times volume \\ = 10^{- pH} M \times volume \\ = 10^{- 4.0} M \times 0.100 L \\ = 1.0 \times 10^{- 5} {mol H}^{+} \end{aligned}$ ‍

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

The molarity of the diluted solution can be calculated by using the moles of

H^{+}

from the original solution and the total volume after dilution.

\begin{aligned} [H^{+}]_{f i n a l} & = \frac{{mol H}^{+}}{L solution} \\ = \frac{1.0 \times 10^{- 5} {mol H}^{+}}{1.0 L} \\ = 1.0 \times 10^{- 5} M \end{aligned}

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

Finally, we can use Eq. 1a to calculate

pH

\begin{aligned} pH & = - \log [H^{+}] \\ = - log (1.0 \times 10^{- 5}) \\ = 5.0 \end{aligned}

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

pH

. 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

pH

changes.

Relationship between $pH$ ‍ and acid strength

Based on the equation for

pH

, we know that

pH

is related to

[H^{+}]

. However, it is important to remember that

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

[H^{+}]

at a given acid concentration. For example, a

1.0 M

solution of strong acid

HCl

will have a higher concentration of

H^{+}

than a

1.0 M

solution of weak acid

HF

. Thus, for two solutions of monoprotic acid at the same concentration,

pH

will be proportional to acid strength.

More generally though, both acid strength and concentration determine

[H^{+}]

. Therefore, we can't always assume that the

pH

of a strong acid solution will be lower than the

pH

of a weak acid solution. The acid concentration matters too!

Summary

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

$\begin{aligned} pH & = - \log [H^{+}] \\ [H^{+}] & = 10^{- pH} \end{aligned}$ ‍

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

$\begin{aligned} pOH & = - \log [{OH}^{-}] \\ [{OH}^{-}] & = 10^{- pOH} \end{aligned}$ ‍

For every factor of $10$ ‍ increase in concentration of $[H^{+}]$ ‍, $pH$ ‍ will decrease by $1$ ‍ unit, and vice versa.
For any aqueous solution at $25^{\circ} C$ ‍:

$pH + pOH = 14$ ‍.

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

Problem 1: Calculating the pH of a strong base solution at $25^{\circ} C$ ‍

We make

200 mL

of a solution with a

0.025 M

concentration of

{Ca(OH)}_{2}

. The solution is then diluted to

1.00 L

by adding additional water.

What is the $pH$ ‍ of the solution after dilution?

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Jonathan Ziesmer
Posted 8 years ago. Direct link to Jonathan Ziesmer's post “How does the temperature ...”
How does the temperature affect the pH and pOH?
Button navigates to signup pageComment on Jonathan Ziesmer's post “How does the temperature ...”
(37 votes)
Answer
- Matt B
  Posted 8 years ago. Direct link to Matt B's post “At 100C the pH of water i...”
  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.
  Comment on Matt B's post “At 100C the pH of water i...”
  (44 votes)
Kaylee Wilson
Posted 8 years ago. Direct link to Kaylee Wilson's post “What does M stand for in ...”
What does M stand for in the unit labels?
Button navigates to signup pageComment on Kaylee Wilson's post “What does M stand for in ...”
(15 votes)
Answer
- SRIVATHSAN B
  Posted 8 years ago. Direct link to SRIVATHSAN B's post “M stands for the unit of ...”
  M stands for the unit of Molarity of the solution.
  Comment on SRIVATHSAN B's post “M stands for the unit of ...”
  (20 votes)
Amit Mukherjee
Posted 8 years ago. Direct link to Amit Mukherjee's post “Could someone explain the...”
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?
Button navigates to signup pageButton navigates to signup page
(8 votes)
Answer
- Matt B
  Posted 8 years ago. Direct link to Matt B's post “Nice question!! It is imp...”
  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.
  Comment on Matt B's post “Nice question!! It is imp...”
  (25 votes)
SRIVATHSAN B
Posted 8 years ago. Direct link to SRIVATHSAN B's post “Can someone please explai...”
Can someone please explain what are monoprotic and diprotic acids? Thanks.
Button navigates to signup pageButton navigates to signup page
(13 votes)
Answer
- Davide Ghazal
  Posted 8 years ago. Direct link to Davide Ghazal's post “H2SO4 is a typical diprot...”
  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+
  Comment on Davide Ghazal's post “H2SO4 is a typical diprot...”
  (16 votes)
areebaali160
Posted 8 years ago. Direct link to areebaali160's post “how can we solve pH,pOH n...”
how can we solve pH,pOH numericals without using scientific calculator during our examination?
Button navigates to signup pageButton navigates to signup page
(8 votes)
Answer
- Matt B
  Posted 8 years ago. Direct link to Matt B's post “Unless you are using perf...”
  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
  Comment on Matt B's post “Unless you are using perf...”
  (7 votes)
apuri
Posted 8 years ago. Direct link to apuri's post “How can NaOH have a pH sc...”
How can NaOH have a pH scale? How can a base add H+ ions to the solution? It adds OH- ions right?
Button navigates to signup pageButton navigates to signup page
(5 votes)
Answer
- Ernest Zinck
  Posted 8 years ago. Direct link to Ernest Zinck's post “NaOH does not have a pH, ...”
  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.
  Comment on Ernest Zinck's post “NaOH does not have a pH, ...”
  (11 votes)
Michelle
Posted 8 years ago. Direct link to Michelle's post “How does pH+pOH= 14? Wher...”
How does pH+pOH= 14? Where does the random number come in?
Button navigates to signup pageButton navigates to signup page
(3 votes)
Answer
- Ernest Zinck
  Posted 8 years ago. Direct link to Ernest Zinck's post “There is no random number...”
  There is no random number.
  The formula comes from the ion product for water.
  [H⁺][OH⁻] = 1.00 × 10⁻¹⁴
  ∴ pH + pOH = 14.00
  Comment on Ernest Zinck's post “There is no random number...”
  (4 votes)
m pe
Posted 6 years ago. Direct link to m pe's post “Hi I feel like there's a ...”
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!
Button navigates to signup pageButton navigates to signup page
(3 votes)
Answer
- RogerP
  Posted 6 years ago. Direct link to RogerP's post “This is the _power rule_ ...”
  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.
  Comment on RogerP's post “This is the _power rule_ ...”
  (3 votes)
Nicolas Jaramillo
Posted 7 years ago. Direct link to Nicolas Jaramillo's post “what is -log? is it a num...”
what is -log? is it a number?
Button navigates to signup pageButton navigates to signup page
(2 votes)
Answer
- Ryan W
  Posted 7 years ago. Direct link to Ryan W's post “Log is a maths function, ...”
  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.
  Button navigates to signup page
  (4 votes)
Aditya Shelke
Posted 6 years ago. Direct link to Aditya Shelke's post “Is it necessary that to c...”
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 :)
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(3 votes)
Answer
- Shlok Sanju
  Posted 5 months ago. Direct link to Shlok Sanju's post “Ok so while calculating t...”
  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!
  Button navigates to signup page
  (1 vote)

Course: Chemistry library > Unit 11

pH, pOH, and the pH scale

Key points

Introduction

Definitions of $pH$ ‍ and $pOH$ ‍

Relating $[H^{+}]$ ‍ and $pH$ ‍

Relating $[{OH}^{-}]$ ‍ and $pOH$ ‍

Relating $pH$ ‍ and $pOH$ ‍

Example 1: Calculating the $pH$ ‍ of a strong base solution

Step 1. Calculate the molar concentration of $NaOH$ ‍

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

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

Step 4: Calculate $pH$ ‍ from $pOH$ ‍ using Eq. 3

The $pH$ ‍ scale: Acidic, basic, and neutral solutions

Example $2$ ‍: Determining the $pH$ ‍ of a diluted strong acid solution

Method 1. Use properties of the log scale

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

Step 1: Calculate moles of $H^{+}$ ‍

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

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

Relationship between $pH$ ‍ and acid strength

Summary

Attributions

Additional References

Problem 1: Calculating the pH of a strong base solution at $25^{\circ} C$ ‍

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Chemistry library

Course: Chemistry library > Unit 11

Key points

Introduction

Definitions of pH‍ and pOH‍

Relating [H+]‍ and pH‍

Relating [OH−]‍ and pOH‍

Relating pH‍ and pOH‍

Example 1: Calculating the pH‍ of a strong base solution

Step 1. Calculate the molar concentration of NaOH‍

Step 2: Calculate [OH−]‍ based on the dissociation of NaOH‍

Step 3: Calculate pOH‍ from [OH−]‍ using Eq. 2a

Step 4: Calculate pH‍ from pOH‍ using Eq. 3

The pH‍ scale: Acidic, basic, and neutral solutions

Example 2‍ : Determining the pH‍ of a diluted strong acid solution

Method 1. Use properties of the log scale

Method 2. Use moles of H+‍ to calculate pH‍

Step 1: Calculate moles of H+‍

Step 2: Calculate molarity of H+‍ after dilution

Step 3: Calculate pH‍ from [H+]‍

Relationship between pH‍ and acid strength

Summary

Problem 1: Calculating the pH of a strong base solution at 25∘C‍

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Definitions of $pH$ ‍ and $pOH$ ‍

Relating $[H^{+}]$ ‍ and $pH$ ‍

Relating $[{OH}^{-}]$ ‍ and $pOH$ ‍

Relating $pH$ ‍ and $pOH$ ‍

Example 1: Calculating the $pH$ ‍ of a strong base solution

Step 1. Calculate the molar concentration of $NaOH$ ‍

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

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

Step 4: Calculate $pH$ ‍ from $pOH$ ‍ using Eq. 3

The $pH$ ‍ scale: Acidic, basic, and neutral solutions

Example $2$ ‍: Determining the $pH$ ‍ of a diluted strong acid solution

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

Step 1: Calculate moles of $H^{+}$ ‍

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

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

Relationship between $pH$ ‍ and acid strength

Problem 1: Calculating the pH of a strong base solution at $25^{\circ} C$ ‍