14.2 pH & pOH

In the previous section, the relationship between the OH ^–, H ⁺ (or accurately H₃O ⁺) and Kw is explained.

Kw = [ H₃O ⁺ ] x [ OH ^– ] = 1.0 x 10  ^-14

Taking the negative logarithm of both sides ( – log ):

-log Kw = – log { [ H₃O ⁺ ] x [ OH ^– ] } =  – log (1.0 x 10  ^-14 )

Using the p – function: p X = – log X, then it follows that:

The formula above is abbreviated to:

pH   +   POH   = 14

where pH = – log H and pOH = – log OH

The above relationships are interconnected:

and the [H ⁺] can be calculated using the anti-logarithm: [H ⁺] = 10 ^{– pH}

Similarly,

[OH ^–] =  10  ^{– pOH}

The pH is therefore defined the tool to express and measure the acidity of an aqueous solution. The pH scale expresses the range of the acidity and the basicity of an aqueous solution:

At pH = 0 – 6, the aqueous solution is said to acidic and the [ H ⁺ ] is greater than 1.0 x 10 ^-7 M

At pH = 7, the aqueous solution is said to be neutral and the [ H ⁺ ] = [ OH  ^– ] = 1.0 x 10 ^-7 M

At pH = 8 – 14, the aqueous solution is said to be basic and [ H ⁺ ] is smaller than 1.0 x 10 ^-7 M

Similarly,

At pOH = 0 – 6, the aqueous solution is said to basic and the [ H ⁺ ] is less than 1.0 x 10 ^-7 M

At pOH = 7, the aqueous solution is said to be neutral and the [ H ⁺ ] = [ OH  ^– ] = 1.0 x 10 ^-7 M

At pOH = 8 – 14, the aqueous solution is said to be acidic and [ H ⁺ ] is greater than 1.0 x 10 ^-7 M

The diagram below exhibits some aqueous solutions at different pH values: It also exhibits the relationships between H ⁺, OH ^–, pH and pOH. It also shows  relative strengths of conjugate acid-base pairs, as indicated by their ionization constants in aqueous solution.

Reference: https://chem.libretexts.org/Bookshelves/General_Chemistry/Map%3A_Principles_of_Modern_Chemistry_(Oxtoby_et_al.)/Unit_4%3A_Equilibrium_in_Chemical_Reactions/15%3A_AcidBase_Equilibria/15.3%3A_Acid_and_Base_Strength

The diagram above can be drawn as a logarithmic graph:

Reference:

https://chemed.chem.purdue.edu/genchem/topicreview/bp/ch17/ph.php

pH and pOH Calculations

Example:

A Drano solution has a solution has a concentration of sodium hydroxide (NaOH) of 2.86 M, calculate the OH ^–, H ⁺ and determine the pH of the solution.

NaOH is strong base:

NaOH will dissociate completely in aqueous solution giving Na ⁺ and OH ^–

Therefore, the [ OH ^– ] = 2.86 M

[ H ⁺ ]  = [ Kw ] / [ [ OH ^– ] = [ 1.0 x 10  ^-14 ] / [ 2.86 ] = 3.50 x 10  ^-15

The pH = – log [ H ⁺ ] =  – log [ 3.50 x 10  ^-15 ] = 14.5

Example:

Calculate the pH of 2.50 M HCl.

HCl is strong acid and therefore it dissociates completely in aqueous solution giving H ⁺ and Cl ^–.

[ H ⁺ ] = 2.50 M

pH = – log 2.50 = – 0.398

pH and Significant Figures

General Rules:

1. A whole-number pH corresponds to a concentration known only as a power of ten.
2. A pH to one decimal place (like 3.8) corresponds to a concentration known to one significant figure.
3. A pH to two decimal places (like 3.89) corresponds to a concentration known to two significant figures.
4. The part of the pH to the left of the decimal point has no effect on significant figures.

Examples:

Calculate the pH of a solution where the [H⁺] is 0.00500 M.

The value 0.00500 has 3 significant figures, and therefore, the pH has to reflect 3 significant figures

pH = – log [ 0.00500 ] = 2.301029995663981

The whole number (i.e. 2) left to the decimal point has no effect

The answer is rounded off to the correct significant figures and it has to have 3 significant figures after the decimal point:

pH = 2.301

Calculate the pH of a solution that has 1.3874 M

The value 1.3874 has 5 significant figures, and therefore, the pH has to reflect 5 significant figures

pH = – log [ 1.3874 ] = 0.1422016901635133

The whole number (i.e. 1) left to the decimal point has no effect.

The answer is rounded off to the correct significant figures and it has to have 5 significant figures after the decimal point:

pH = 0. 14220

The whole number (i.e. 0) left to the decimal point has no effect.

What is the pOH of a 0.0235 M HCI solution?

pH = -log[H+] = -log [ 0.0235 ] = 1.629

pOH = 14.000 – pH = 14.000 – 1.629 = 12.371

What is the pH of a 6.50 x 10^-3 M KOH solution?

pOH = -log [ OH ^– ] = -log [ 6.50 x 10 ^-3 ]  = 2.187

pH = 14.000 – pOH = 14.000 – 2.187 = 11.813

Similarly if the pH is known, then the concentration of [ H ⁺ ] and [ OH ^– ].

Example:

What is the concentration of [ H ⁺] and [ OH ^– ] of HCl solution if it has a pH of 3.7561

We have to ignore the whole number left of the decimal point (i.e. 3).

We consider only the digits after the decimal point. They are 4 digits. Hence the concentration of [ H ⁺] and [ OH ^– ] of HCl will have 4 significant figures.

pH = – log [ H ⁺] = 3.7561

[ H ⁺]  = 10  ^{– pH} = 10 ^{– 3.7561} = 1.753476702422828 x 10 ^{– 4} = 1.753 x 10 ^{– 4} M

[ OH ^– ] = [ 1.000 x 10 ^-14 ]  /  [ 1.753 x 10  ^-4 ] = 5.704506560183 x 10 ^-11 = 5.704 x 10 ^-11 M

Experimentally is the pH is measured and the pOH is calculated.

The pH is measured experimentally by using the pH meters. pH meters come in different styles but the concepts in all of them are the same. Good pH meters can measure the concentration of the H ⁺ with accuracy of +/- 0.002

Much more accurate pH meters are called Saturated Calomel Electrode (SEC) which is based on mercury – I – chloride Hg₂Cl₂ and Ag/AgCl Electrode

Reference: https://chem.libretexts.org/Bookshelves/Analytical_Chemistry/Supplemental_Modules_(Analytical_Chemistry)/Analytical_Sciences_Digital_Library/JASDL/Courseware/Analytical_Electrochemistry%3A_Potentiometry/03_Potentiometric_Theory/04_Reference_Electrodes

In case of Saturated Calomel Electrode, a paste is prepared from solution that is saturated with KCl.

The measurement cell is made through a porous glass frit or fiber which allows the movement of ions, but will not allow the bulk solution. There are several electrodes designed for potentiometry where the reference is a half cell, which is part of sensing structure of the electrode. Such set up is referred to as a combination electrode.

The pH, pOH and Kw definition and calculations are illustrated in the videos below:

The Phet pH simulation

https://phet.colorado.edu/sims/html/ph-scale/latest/ph-scale_en.html

pH Scale Guided Inquiry Activity

            You’re probably familiar with the sour taste of acidic lemon juice and the slippery feel of alkaline (basic) soap. In fact, these characteristics were used to identify acids from bases long ago. Today, we understand much more about acidity and alkalinity, far beyond taste and feel.

            Acids are most simply defined as proton donors. They are substances that react with water to produce H₃O⁺ (hydronium) ions (after all, H₂O + one proton → H₃O⁺). When you add an acid to water, the water acts as a base, accepting protons from the acid.

H₂O + one proton → H₃O⁺

Bases are most simply defined as proton acceptors. They react with water to produce OH^– (hydroxide) ions (H₂O – one proton → OH^–). When you add a base to water, the water acts as an acid, donating protons to the base.

            H₂O – one proton → OH^–

            Some acids and bases ionize only partially in water, while others ionize almost entirely. To quantify how many molecules of an acid or base ionize in water (which we call the strength of the solution), we use the pH scale. A solution’s place on the pH scale illustrates how many hydronium (H₃O⁺) ions are present in that solution. The pH scale is centered around 7, because water contains 1 x 10^–7 moles of H₃O⁺ ions per liter. A solution with 100x more H₃O⁺ ions than water has 1 x 10^–5 moles of H₃O⁺ per liter, which gives it a pH of 5. This would be considered a weak acid. A solution with 1,000,000x more H₃O⁺ than water has 1 x 10^–1 moles of H₃O⁺ ions per liter. Any such substance has a pH of 1 and is considered a strong acid. On the other side, a solution with 1/100 the H₃O⁺ ion concentration of water has 1 x 10^–9 moles of H₃O⁺ ions per liter. This solution has a pH of 9 and is considered a weak base.

            This complicated logarithmic calculation gives us a simple 0-14 pH scale:

                         Strong acids … Weak acids             Weak bases … Strong bases

                          lots of H₃O⁺  …  some H₃O⁺                        little H₃O⁺   … very little H₃O⁺

Procedure:

1. Go to phet.colorado.edu.
2. Click on the Chemistry tab.
3. Choose the pH Scale simulation: https://phet.colorado.edu/en/simulation/ph-scale
4. Click on Run in HTML5, then choose the Micro tab.
5. Add battery acid, then use the bottom faucet to drain the tank until 0.1 L remains.

1) Record the pH of the solution:

2) How many moles of H₃O⁺ (hydronium) ions are present in each liter of the solution? ***Answer in scientific notation and decimal form.

Recall how pH relates to the exponent of the concentration of H₃O⁺ ions.

3) How many moles of H₃O⁺ ions are present in 0.1 L of the solution (the volume in the cup)?

• Add water until there is 1 L of solution in the container. Compared to the H₃O⁺ in the battery acid, the H₃O⁺ in the water is negligible. So, we can consider the amount of H₃O⁺ in the diluted solution to be unchanged by the added water.

4) Since the amount of hydronium is essentially unchanged, how many moles of H₃O⁺ ions are present in the liter of diluted solution?

5) What is the new concentration of H₃O⁺ ions, in moles per liter?

6) What is the pH of the new diluted solution?

7) Explain why diluting the battery acid increased the pH by 1.

• Drain the 10% battery acid solution until 0.1 L remains. Be careful not to drain too much of the solution. If you drain too much, you will have to restart the experiment from the beginning.
• Pour water into the container until the volume again reaches 1 liter. Be careful not to pour too much water into the solution.

8) After again increasing the volume by a factor of 10, what is the new concentration of H₃O⁺ ions?

9) What is the pH of the new solution?

I. Drain the 1% battery acid solution until 0.1 L remains.

J. Pour water into the container until the volume again reaches 1 liter.

10) After again increasing the volume by a factor of 10, what is the new concentration of H₃O⁺ ions?

11) What is the pH of the new solution?

K. Drain the 0.1% battery acid solution until 0.1 L remains.

L. Pour water into the container until the volume again reaches 1 liter.

12) After again increasing the volume by a factor of 10, what is the new concentration of H₃O⁺ ions?

13) What is the pH of the new solution?

13. Drain the 0.01% battery acid solution until 0.1 L remains.

N. Pour water into the container until the volume again reaches 1 liter.

14) After again increasing the volume by a factor of 10, what is the new concentration of H₃O⁺ ions?

15) What is the pH of the new solution?

15. Drain the 0.001% battery acid solution until 0.1 L remains.

P. Pour water into the container until the volume again reaches 1 liter.

16) After again increasing the volume by a factor of 10, what is the new concentration of H₃O⁺ ions?

17) What is the pH of the new solution?

18) Why did the pH increase by a lesser amount than previous dilutions? Is the italicized statement in step F still true? What happens to the difference between the solution’s H₃O⁺ ion concentration and water’s H₃O⁺ ion concentration each time you dilute the solution?

17. Use the Custom tab at the bottom to answer the following questions. Manipulate the amount of hydronium and hydroxide in the custom solution.

19) As the H₃O⁺ ion concentration decreases, the pH __________________________.

20) The product of a solution’s H₃O⁺ concentration and its OH^– concentration must always be 1 x 10^-14. So, if a solution’s H₃O⁺ concentration goes down, the solution’s OH^– concentration must go ___________. In other words, if we have less hydronium, we must have __________________ hydroxide