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Chemistry library

Course: Chemistry library > Unit 13

Lesson 1: Acids, bases, and pH
Science
Chemistry library
Acids and bases
Acids, bases, and pH
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pH, pOH, and the pH scale

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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:
pH=−log⁡[H+][H+]=10−pH\begin{aligned}\text{pH}&=-\log[\text{H}^+]\\ \\ [\text H^+]&=10^{-\text{pH}}\end{aligned}
  • 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:
pOH=−log⁡[OH−][OH−]=10−pOH\begin{aligned}\text{pOH}&=-\log[\text{OH}^-]\\ \\ [\text {OH}^-]&=10^{-\text{pOH}}\end{aligned}
  • 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.
Purple-blue hydrangeas next to pinkish purple hydrangeas.
The color of hydrangea flowers can vary depending on the pH of the soil. Blue flowers usually come from acidic soil with a pH less than 6, and pink flowers come from soil with a pH above 6. Photo from WIkimedia Commons, CC BY 2.0
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:
start text, p, H, end text, equals, minus, log, open bracket, start text, H, end text, start superscript, plus, end superscript, close bracket, start text, left parenthesis, E, q, point, space, 1, a, right parenthesis, end text
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:
start text, p, H, end text, equals, minus, log, left parenthesis, 1, times, 10, start superscript, minus, 5, end superscript, right parenthesis, equals, 5, point, 0
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:
open bracket, start text, H, end text, start superscript, plus, end superscript, close bracket, equals, 10, start superscript, minus, start text, p, H, end text, end superscript, start text, left parenthesis, E, q, point, space, 1, b, right parenthesis, end text

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:
start text, p, O, H, end text, equals, minus, log, open bracket, start text, O, H, end text, start superscript, minus, end superscript, close bracket, space, start text, left parenthesis, E, q, point, space, 2, a, right parenthesis, end text
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:
start text, p, O, H, end text, equals, minus, log, left parenthesis, 1, times, 10, start superscript, minus, 12, end superscript, right parenthesis, equals, 12, point, 0
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:
10, start superscript, minus, start text, p, O, H, end text, end superscript, equals, open bracket, start text, O, H, end text, start superscript, minus, end superscript, close bracket, start text, left parenthesis, E, q, point, space, 2, b, right parenthesis, end text

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:
start text, p, H, end text, plus, start text, p, O, H, end text, equals, 14, space, space, start text, left parenthesis, E, q, point, space, 3, right parenthesis, 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:
start text, M, o, l, a, r, space, c, o, n, c, e, n, t, r, a, t, i, o, n, end text, equals, start fraction, start text, m, o, l, space, s, o, l, u, t, e, end text, divided by, start text, L, space, s, o, l, u, t, i, o, n, end text, end fraction
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:
[NaOH]=1.0 mmol NaOH1.0 L=1.0×10−3 mol NaOH1.0 L=1.0×10−3 M NaOH\begin{aligned}\text{[NaOH]}&=\dfrac{1.0\text{ mmol NaOH}}{1.0\text{ L}}\\ \\ &=\dfrac{1.0\times10^{-3}\text{ mol NaOH}}{1.0\text{ L}}\\ \\ &=1.0\times10^{-3}\text{ M NaOH}\end{aligned}
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:
start text, N, a, O, H, end text, left parenthesis, a, q, right parenthesis, right arrow, start text, N, a, end text, start superscript, plus, end superscript, left parenthesis, a, q, right parenthesis, plus, start text, O, H, end text, start superscript, minus, end superscript, left parenthesis, a, q, right parenthesis
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:
open bracket, start text, N, a, O, H, end text, close bracket, equals, open bracket, start text, O, H, end text, start superscript, minus, end superscript, close bracket, equals, 1, point, 0, times, 10, start superscript, minus, 3, end superscript, start text, space, M, end text

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:
pOH=−log⁡[OH−]=−log⁡(1.0×10−3)=3.00\begin{aligned}\text{pOH}&=-\log[\text{OH}^-]\\ \\ &=-\log(1.0\times10^{-3})\\ \\ &=3.00\end{aligned}
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:
start text, p, H, end text, equals, 14, minus, start text, p, O, 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:
start text, p, H, end text, equals, 14, minus, 3, point, 00, equals, 11, point, 00
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.
The pH scale from -1 to 14.
The pH scale. Acidic solutions have pH values less than 7, and basic solutions have pH values greater than 7. Image from UCDavis ChemWiki, CC BY-NC-SA 3.0 US.
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:
pH=original pH+1.0=4.0+1.0=5.0\begin{aligned}\text{pH}&=\text{original pH}+1.0 \\ \\ &=4.0+1.0\\ \\ &=5.0\end{aligned}
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.
moles H+=[H+]initial×volume=10−pH M×volume=10−4.0 M×0.100 L=1.0×10−5 mol H+\begin{aligned}\text{moles H}^+&=[\text H^+]_{initial} \times \text{volume}\\ \\ &=10^{-\text{pH}}\,\text M \times \text {volume}\\ \\ &=10^{-4.0}\,\text M \times {0.100\text{ L}}\\ \\ &=1.0 \times 10^{-5}\,\text {mol H}^+\end{aligned}

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.
[H+]final=mol H+L solution=1.0×10−5 mol H+1.0 L=1.0×10−5 M\begin{aligned}[\text{H}^+]_{final}&=\dfrac{\text{mol H}^+}{\text{L solution}}\\ \\ &=\dfrac{1.0 \times 10^{-5}\,\text {mol H}^+}{1.0\,\text{L}}\\ \\ &=1.0 \times 10^{-5}\,\text M\end{aligned}

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:
pH=−log⁡[H+]=−log(1.0×10−5)=5.0\begin{aligned}\text{pH}&=-\log[\text{H}^+]\\ \\ &=-\text{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 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

Hand holding wet pH paper with four stripes (from left to right): orange, green-brown, yellow, and red-orange. The paper is held up for comparison against a reference chart of pH values and colors. The wet paper matches the pH 7 on the reference.
Indicator paper can be used to measure the pH of aqueous solutions. The color of the indicator paper in this picture matches a pH value of 7. Photo from Wikimedia Commons, CC BY-SA 2.5
  • 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:
pH=−log⁡[H+][H+]=10−pH\begin{aligned}\text{pH}&=-\log[\text{H}^+]\\ \\ [\text H^+]&=10^{-\text{pH}}\end{aligned}
  • 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:
pOH=−log⁡[OH−][OH−]=10−pOH\begin{aligned}\text{pOH}&=-\log[\text{OH}^-]\\ \\ [\text {OH}^-]&=10^{-\text{pOH}}\end{aligned}
  • 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?
Choose 1 answer:

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  • old spice man green style avatar for user Jonathan Ziesmer
    Posted 7 years ago. Direct link to Jonathan Ziesmer's post “How does the temperature ...”
    How does the temperature affect the pH and pOH?
    (36 votes)
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  • blobby green style avatar for user Kaylee Wilson
    Posted 7 years ago. Direct link to Kaylee Wilson's post “What does M stand for in ...”
    What does M stand for in the unit labels?
    (15 votes)
    Default Khan Academy avatar avatar for user
  • male robot hal style avatar for user SRIVATHSAN B
    Posted 7 years ago. Direct link to SRIVATHSAN B's post “Can someone please explai...”
    Can someone please explain what are monoprotic and diprotic acids? Thanks.
    (13 votes)
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  • leafers seed style avatar for user Amit Mukherjee
    Posted 7 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?
    (8 votes)
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    • old spice man green style avatar for user Matt B
      Posted 7 years ago. Direct link to Matt B's post “Nice question!! It is imp...”
      Good Answer
      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)
  • leaf green style avatar for user areebaali160
    Posted 7 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?
    (8 votes)
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    • old spice man green style avatar for user Matt B
      Posted 7 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
      (7 votes)
  • ohnoes default style avatar for user apuri
    Posted 7 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?
    (5 votes)
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    • spunky sam blue style avatar for user Ernest Zinck
      Posted 7 years ago. Direct link to Ernest Zinck's post “NaOH does not have a pH, ...”
      Good Answer
      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)
  • female robot grace style avatar for user Michelle
    Posted 7 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?
    (3 votes)
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  • duskpin tree style avatar for user m pe
    Posted 5 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!
    (3 votes)
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    • piceratops seed style avatar for user RogerP
      Posted 5 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.
      (3 votes)
  • leafers tree style avatar for user Nicolas Jaramillo
    Posted 6 years ago. Direct link to Nicolas Jaramillo's post “what is -log? is it a num...”
    what is -log? is it a number?
    (2 votes)
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  • piceratops seedling style avatar for user Aditya Shelke
    Posted 5 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 :)
    (3 votes)
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