Testing samples of a given solution with various indicators allows one to narrow down the range of possible pH values for the solution. We’ll show you how this works using a couple of examples. Using Indicators to Find Approximate pH

In our first example question, we’re given that three separate samples of Solution A are obtained and each sample is tested with a different indicator. Three separate samples of Solution A are obtained and each sample is tested with a different indicator.

The results are shown in the following table, Three separate samples of Solution A are obtained and each sample is tested with a different indicator. The results are as follows:

Methyl orange is yellow, thymol blue is yellow, and methyl red is red. Three separate samples of Solution A are obtained and each sample is tested with a different indicator. The results are as follows: IndicatorColour Methyl orangeyellow Thymol blueyellow Methyl redred

We’ll locate methyl orange, methyl red, and thymol blue on the indicator table. It gives us the pH ranges of the various colours of each indicator. Notice thymol blue appears twice on this table.

We’ll start with methyl orange. It’s line on the indicator table, shown on top here, tells us it changes from red to yellow as the pH increases from 3.2 to 4.4 IndicatorColourpH Range Methyl orangeyellow Thymol blueyellow Methyl redred

So below a pH of 3.2, methyl orange is red… IndicatorColourpH Range Methyl orangeyellow Thymol blueyellow Methyl redred pH RED Methyl Orange 14

Between pH 3.2 and 4.4, it changes from red to orange to yellow as the pH increases. IndicatorColourpH Range Methyl orangeyellow Thymol blueyellow Methyl redred pH ORANGE Methyl Orange 14

And above a pH of 4.4, it is yellow. IndicatorColourpH Range Methyl orangeyellow Thymol blueyellow Methyl redred pH YELLOW Methyl Orange 14

The table shown on the bottom here, tells us methyl orange is yellow in Solution A … IndicatorColourpH Range Methyl orangeyellow Thymol blueyellow Methyl redred pH Methyl Orange 14 YELLOW

So that tells us the pH of Solution A is greater than or equal to 4.4 IndicatorColourpH Range Methyl orangeyellow≥ 4.4 Thymol blueyellow Methyl redred pH Methyl Orange 14 YELLOW

Now we’ll look at thymol blue. It’s lines on the indicator table, shown on top here, tell us it turns yellow at a pH of 2.8 IndicatorColourpH Range Methyl orangeyellow≥ 4.4 Thymol blueyellow Methyl redred pH Thymol blue YELLOW

And stays yellow until the pH increases to 8.0 IndicatorColourpH Range Methyl orangeyellow≥ 4.4 Thymol blueyellow Methyl redred pH Thymol blue YELLOW

Our table at the bottom shows us that thymol blue is yellow in Solution A, IndicatorColourpH Range Methyl orangeyellow≥ 4.4 Thymol blueyellow Methyl redred pH Thymol blue YELLOW

So this tells us the pH is somewhere between 2.8 and 8.0 IndicatorColourpH Range Methyl orangeyellow≥ 4.4 Thymol blueyellow2.8–8.0 Methyl redred pH Thymol blue YELLOW

The line for methyl red on the indicator table, shown on top here, tells us that methyl red is red at pH’s below 4.8 IndicatorColourpH Range Methyl orangeyellow≥ 4.4 Thymol blueyellow2.8–8.0 Methyl redred pH Methyl red 14 RED

Orange between 4.8 and 6.0 IndicatorColourpH Range Methyl orangeyellow≥ 4.4 Thymol blueyellow2.8–8.0 Methyl redred pH Methyl red 14 ORANGE

And yellow at any pH’s above 6.0 IndicatorColourpH Range Methyl orangeyellow≥ 4.4 Thymol blueyellow2.8–8.0 Methyl redred pH Methyl red 14 YELLOW

The table at the bottom shows that methyl red is red in Solution A, IndicatorColourpH Range Methyl orangeyellow≥ 4.4 Thymol blueyellow2.8–8.0 Methyl redred pH Methyl red 14 RED

So this indicator tells us the pH is less than or equal to 4.8 IndicatorColourpH Range Methyl orangeyellow≥ 4.4 Thymol blueyellow2.8–8.0 Methyl redred≤ pH Methyl red 14 RED

Because the pH is greater than or equal to 4.4 IndicatorColourpH Range Methyl orangeyellow≥ 4.4 Thymol blueyellow2.8–8.0 Methyl redred≤ ≤ pH ≤ 4.8

And less than or equal to 4.8, IndicatorColourpH Range Methyl orangeyellow≥ 4.4 Thymol blueyellow2.8–8.0 Methyl redred≤ ≤ pH ≤ 4.8

It means the pH is somewhere between 4.4 and 4.8 inclusive. IndicatorColourpH Range Methyl orangeyellow≥ 4.4 Thymol blueyellow2.8–8.0 Methyl redred≤ 4.8 pH is 4.4 – 4.8

Notice that the range 4.4 to 4.8 is well within the range 2.8 to 8.0 IndicatorColourpH Range Methyl orangeyellow≥ 4.4 Thymol blueyellow2.8–8.0 Methyl redred≤ 4.8 pH is 4.4 – 4.8

So the range thymol blue gives us (click) does not help to narrow down the pH range of 4.4 to 4.8. IndicatorColourpH Range Methyl orangeyellow≥ 4.4 Thymol blueyellow2.8–8.0 Methyl redred≤ 4.8 pH is 4.4 – 4.8

Here’s another question. We’re given that Solution B is yellow when the indicator alizarin yellow is added and blue when the indicator thymol blue is added. Solution B is yellow when alizarin yellow is added and blue when thymol blue is added.

And we’re asked which one of these solutions is the correct identity for solution B. Solution B is yellow when alizarin yellow is added and blue when thymol blue is added. Which of the following correctly identifies Solution B? A.3.2 × 10 –4 M HCl B.3.2 × 10 –4 M NaOH C.6.3 × 10 –5 M HCl D.6.3 × 10 –5 M NaOH

The indicator table tells us that when alizarin yellow is yellow, the pH is less than or equal to Solution B is yellow when alizarin yellow is added and blue when thymol blue is added. Which of the following correctly identifies Solution B? A. 3.2 × 10 –4 M HCl B. 3.2 × 10 –4 M NaOH C. 6.3 × 10 –5 M HCl D. 6.3 × 10 –5 M NaOH IndicatorColourpH Range Alizarin yellowyellow Thymol blueblue pH ≤ 10.1

So we’ll add this range to our table down here for Solution B. Solution B is yellow when alizarin yellow is added and blue when thymol blue is added. Which of the following correctly identifies Solution B? A. 3.2 × 10 –4 M HCl B. 3.2 × 10 –4 M NaOH C. 6.3 × 10 –5 M HCl D. 6.3 × 10 –5 M NaOH IndicatorColourpH Range Alizarin yellowyellow≤ 10.1 Thymol blueblue pH ≤ 10.1

The indicator table also tells us that when thymol blue is blue, the pH is greater than or equal to 9.6. Solution B is yellow when alizarin yellow is added and blue when thymol blue is added. Which of the following correctly identifies Solution B? A. 3.2 × 10 –4 M HCl B. 3.2 × 10 –4 M NaOH C. 6.3 × 10 –5 M HCl D. 6.3 × 10 –5 M NaOH IndicatorColourpH Range Alizarin yellowyellow≤ 10.1 Thymol blueblue pH ≤ 10.1pH ≥ 9.6

So we’ll add this range in the table down here. Solution B is yellow when alizarin yellow is added and blue when thymol blue is added. Which of the following correctly identifies Solution B? A. 3.2 × 10 –4 M HCl B. 3.2 × 10 –4 M NaOH C. 6.3 × 10 –5 M HCl D. 6.3 × 10 –5 M NaOH IndicatorColourpH Range Alizarin yellowyellow≤ 10.1 Thymol blueblue≥ 9.6 pH ≤ 10.1pH ≥ 9.6

Because the pH is greater than or equal to 9.6 and less than or equal to 10.1, Solution B is yellow when alizarin yellow is added and blue when thymol blue is added. Which of the following correctly identifies Solution B? A. 3.2 × 10 –4 M HCl B. 3.2 × 10 –4 M NaOH C. 6.3 × 10 –5 M HCl D. 6.3 × 10 –5 M NaOH IndicatorColourpH Range Alizarin yellowyellow≤ 10.1 Thymol blueblue≥ 9.6 pH ≤ 10.1 pH ≥ 9.6

We can say that the pH is somewhere between 9.6 and 10.1, inclusive. Solution B is yellow when alizarin yellow is added and blue when thymol blue is added. Which of the following correctly identifies Solution B? A. 3.2 × 10 –4 M HCl B. 3.2 × 10 –4 M NaOH C. 6.3 × 10 –5 M HCl D. 6.3 × 10 –5 M NaOH IndicatorColourpH Range Alizarin yellowyellow≤ 10.1 Thymol blueblue≥ 9.6 pH ≤ 10.1 pH ≥ 9.6 pH is between 9.6–10.1

Now, we’ll calculate the pH of each one of the given possible solutions and see which one fits within with the range we came up with. We’ll start with 3.2 × 10 –4 M HCl Solution B is yellow when alizarin yellow is added and blue when thymol blue is added. Which of the following correctly identifies Solution B? A. 3.2 × 10 –4 M HCl B. 3.2 × 10 –4 M NaOH C. 6.3 × 10 –5 M HCl D. 6.3 × 10 –5 M NaOH

Because HCl is a strong acid, the concentration of H3O + is equal to the concentration of the acid, which is 3.2 × 10 –4 M. Solution B is yellow when alizarin yellow is added and blue when thymol blue is added. Which of the following correctly identifies Solution B? A. 3.2 × 10 –4 M HCl B. 3.2 × 10 –4 M NaOH C. 6.3 × 10 –5 M HCl D. 6.3 × 10 –5 M NaOH

pH is the negative log of the hydronium ion concentration Solution B is yellow when alizarin yellow is added and blue when thymol blue is added. Which of the following correctly identifies Solution B? A. 3.2 × 10 –4 M HCl B. 3.2 × 10 –4 M NaOH C. 6.3 × 10 –5 M HCl D. 6.3 × 10 –5 M NaOH

Which is the negative log of 3.2 × 10 –4. Solution B is yellow when alizarin yellow is added and blue when thymol blue is added. Which of the following correctly identifies Solution B? A. 3.2 × 10 –4 M HCl B. 3.2 × 10 –4 M NaOH C. 6.3 × 10 –5 M HCl D. 6.3 × 10 –5 M NaOH

Rounded to one decimal place, this comes out to 3.5 Solution B is yellow when alizarin yellow is added and blue when thymol blue is added. Which of the following correctly identifies Solution B? A. 3.2 × 10 –4 M HCl B. 3.2 × 10 –4 M NaOH C. 6.3 × 10 –5 M HCl D. 6.3 × 10 –5 M NaOH

So we’ll make a note of that here. Solution B is yellow when alizarin yellow is added and blue when thymol blue is added. Which of the following correctly identifies Solution B? A. 3.2 × 10 –4 M HCl B. 3.2 × 10 –4 M NaOH C. 6.3 × 10 –5 M HCl D. 6.3 × 10 –5 M NaOH

Now, we’ll find the pH of the second solution, 3.2 × 10 –4 M NaOH. Solution B is yellow when alizarin yellow is added and blue when thymol blue is added. Which of the following correctly identifies Solution B? A. 3.2 × 10 –4 M HCl B. 3.2 × 10 –4 M NaOH C. 6.3 × 10 –5 M HCl D. 6.3 × 10 –5 M NaOH

Because NaOH is a strong base, we can say the concentration of OH minus is equal to the concentration of NaOH, which is 3.2 × 10 –4 M. Solution B is yellow when alizarin yellow is added and blue when thymol blue is added. Which of the following correctly identifies Solution B? A. 3.2 × 10 –4 M HCl B. 3.2 × 10 –4 M NaOH C. 6.3 × 10 –5 M HCl D. 6.3 × 10 –5 M NaOH

pOH is the negative log of the hydroxide ion concentration Solution B is yellow when alizarin yellow is added and blue when thymol blue is added. Which of the following correctly identifies Solution B? A. 3.2 × 10 –4 M HCl B. 3.2 × 10 –4 M NaOH C. 6.3 × 10 –5 M HCl D. 6.3 × 10 –5 M NaOH

Which is the negative log of 3.2 × 10 –4. Solution B is yellow when alizarin yellow is added and blue when thymol blue is added. Which of the following correctly identifies Solution B? A. 3.2 × 10 –4 M HCl B. 3.2 × 10 –4 M NaOH C. 6.3 × 10 –5 M HCl D. 6.3 × 10 –5 M NaOH

So the pOH, expressed to one decimal place is 3.5. Solution B is yellow when alizarin yellow is added and blue when thymol blue is added. Which of the following correctly identifies Solution B? A. 3.2 × 10 –4 M HCl B. 3.2 × 10 –4 M NaOH C. 6.3 × 10 –5 M HCl D. 6.3 × 10 –5 M NaOH

The pH is 14 minus the pOH or 14 minus 3.5, which comes out to 10.5 Solution B is yellow when alizarin yellow is added and blue when thymol blue is added. Which of the following correctly identifies Solution B? A. 3.2 × 10 –4 M HCl B. 3.2 × 10 –4 M NaOH C. 6.3 × 10 –5 M HCl D. 6.3 × 10 –5 M NaOH

Which we’ll make a note of here. Solution B is yellow when alizarin yellow is added and blue when thymol blue is added. Which of the following correctly identifies Solution B? A. 3.2 × 10 –4 M HCl B. 3.2 × 10 –4 M NaOH C. 6.3 × 10 –5 M HCl D. 6.3 × 10 –5 M NaOH

Now, we’ll consider the third solution, 6.3 × 10 –5 M HCl. Solution B is yellow when alizarin yellow is added and blue when thymol blue is added. Which of the following correctly identifies Solution B? A. 3.2 × 10 –4 M HCl B. 3.2 × 10 –4 M NaOH C. 6.3 × 10 –5 M HCl D. 6.3 × 10 –5 M NaOH

Again, because HCl is a strong acid, the hydronium ion concentration is equal to the concentration of the acid, which is 6.3 × 10 –5 M. Solution B is yellow when alizarin yellow is added and blue when thymol blue is added. Which of the following correctly identifies Solution B? A. 3.2 × 10 –4 M HCl B. 3.2 × 10 –4 M NaOH C. 6.3 × 10 –5 M HCl D. 6.3 × 10 –5 M NaOH

pH is the negative log of the hydronium ion concentration, Solution B is yellow when alizarin yellow is added and blue when thymol blue is added. Which of the following correctly identifies Solution B? A. 3.2 × 10 –4 M HCl B. 3.2 × 10 –4 M NaOH C. 6.3 × 10 –5 M HCl D. 6.3 × 10 –5 M NaOH

or the negative log of 6.3 × 10 –5. Solution B is yellow when alizarin yellow is added and blue when thymol blue is added. Which of the following correctly identifies Solution B? A. 3.2 × 10 –4 M HCl B. 3.2 × 10 –4 M NaOH C. 6.3 × 10 –5 M HCl D. 6.3 × 10 –5 M NaOH

Which, to one decimal place, comes out to 4.2 Solution B is yellow when alizarin yellow is added and blue when thymol blue is added. Which of the following correctly identifies Solution B? A. 3.2 × 10 –4 M HCl B. 3.2 × 10 –4 M NaOH C. 6.3 × 10 –5 M HCl D. 6.3 × 10 –5 M NaOH

Which we’ll make a note of here. Solution B is yellow when alizarin yellow is added and blue when thymol blue is added. Which of the following correctly identifies Solution B? A. 3.2 × 10 –4 M HCl B. 3.2 × 10 –4 M NaOH C. 6.3 × 10 –5 M HCl D. 6.3 × 10 –5 M NaOH

Now we’ll find the pH of the last solution, 6.3 × 10 –5 M NaOH. Solution B is yellow when alizarin yellow is added and blue when thymol blue is added. Which of the following correctly identifies Solution B? A. 3.2 × 10 –4 M HCl B. 3.2 × 10 –4 M NaOH C. 6.3 × 10 –5 M HCl D. 6.3 × 10 –5 M NaOH

Because this is a strong base, the hydroxide ion concentration is equal to the concentration of NaOH, which is 6.3 × 10 –5 M. Solution B is yellow when alizarin yellow is added and blue when thymol blue is added. Which of the following correctly identifies Solution B? A. 3.2 × 10 –4 M HCl B. 3.2 × 10 –4 M NaOH C. 6.3 × 10 –5 M HCl D. 6.3 × 10 –5 M NaOH

pOH is the negative log of the hydroxide ion concentration Solution B is yellow when alizarin yellow is added and blue when thymol blue is added. Which of the following correctly identifies Solution B? A. 3.2 × 10 –4 M HCl B. 3.2 × 10 –4 M NaOH C. 6.3 × 10 –5 M HCl D. 6.3 × 10 –5 M NaOH

Which is the negative log of 6.3 × 10 –5. Solution B is yellow when alizarin yellow is added and blue when thymol blue is added. Which of the following correctly identifies Solution B? A. 3.2 × 10 –4 M HCl B. 3.2 × 10 –4 M NaOH C. 6.3 × 10 –5 M HCl D. 6.3 × 10 –5 M NaOH

Rounded to 1 decimal place, the pOH comes out to 4.2 Solution B is yellow when alizarin yellow is added and blue when thymol blue is added. Which of the following correctly identifies Solution B? A. 3.2 × 10 –4 M HCl B. 3.2 × 10 –4 M NaOH C. 6.3 × 10 –5 M HCl D. 6.3 × 10 –5 M NaOH

The pH is 14 minus the pOH or 14 minus 4.2, which comes out to 9.8. Solution B is yellow when alizarin yellow is added and blue when thymol blue is added. Which of the following correctly identifies Solution B? A. 3.2 × 10 –4 M HCl B. 3.2 × 10 –4 M NaOH C. 6.3 × 10 –5 M HCl D. 6.3 × 10 –5 M NaOH

So we’ll make a note of that here. Solution B is yellow when alizarin yellow is added and blue when thymol blue is added. Which of the following correctly identifies Solution B? A. 3.2 × 10 –4 M HCl B. 3.2 × 10 –4 M NaOH C. 6.3 × 10 –5 M HCl D. 6.3 × 10 –5 M NaOH

Remember, the tests with the indicators told us that the pH of Solution B is somewhere between 9.6 and Solution B is yellow when alizarin yellow is added and blue when thymol blue is added. Which of the following correctly identifies Solution B? A. 3.2 × 10 –4 M HCl B. 3.2 × 10 –4 M NaOH C. 6.3 × 10 –5 M HCl D. 6.3 × 10 –5 M NaOH

We see that the only pH which fits inside this range is 9.8. Solution B is yellow when alizarin yellow is added and blue when thymol blue is added. Which of the following correctly identifies Solution B? A. 3.2 × 10 –4 M HCl B. 3.2 × 10 –4 M NaOH C. 6.3 × 10 –5 M HCl D. 6.3 × 10 –5 M NaOH

And the solution with a pH of 9.8 is choice D, 6.3 × 10 –5 M NaOH. Solution B is yellow when alizarin yellow is added and blue when thymol blue is added. Which of the following correctly identifies Solution B? A. 3.2 × 10 –4 M HCl B. 3.2 × 10 –4 M NaOH C. 6.3 × 10 –5 M HCl D. 6.3 × 10 –5 M NaOH

So to summarize, we can say that using the results of the indicators, we can conclude that Solution B is 6.3 × 10 –5 M NaOH. Solution B is yellow when alizarin yellow is added and blue when thymol blue is added. Which of the following correctly identifies Solution B? A. 3.2 × 10 –4 M HCl B. 3.2 × 10 –4 M NaOH C. 6.3 × 10 –5 M HCl D. 6.3 × 10 –5 M NaOH Using the results of the indicators, we can conclude that Solution B is 6.3 × 10 –5 M NaOH