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pH and pKa

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If you have ever tried lemon juice, then you and I can agree that lemon juice has a very acidic taste. Lemon juice is a type of **weak acid**, and to learn about the **pH **and **pK**_{a }of weak acids, we need to dive into the world of K_{a}, ICE tables and even percent ionization!

- First, we will talk about definitions of pH and pKa
- Then, we will look at calculations involving pH and pKa
- Lastly, we will learn about percent ionization.

Before diving into pH and pKa, let's recall the definition of Bronsted-Lowry acids and bases, and also the meaning of conjugate acids and bases.

**Bronsted-Lowry acids** are proton (H^{+}) donors, whereas **Bronsted-Lowry bases** are proton (H^{+}) acceptors. Let's look at the reaction between ammonia and water.

_{The reaction between ammonia and water, Isadora Santos - StudySmarter Originals.}

**Conjugate acids** are *bases *that gained a proton H^{+}. On the other hand, **Conjugate bases** are *acids *that lost a proton H^{+}. For example, when HCl is added to H_{2}O, it dissociates to form H3O^{+ }and Cl^{-}. Water will gain a proton, and HCl will lose a proton.

_{Conjugate pairs in a reaction between HCl and Water, Isadora Santos - StudySmarter Originals.}

Some chemistry books use H^{+ }instead of ^{ }H3O^{+} to refer to hydrogen ions. However, these two terms can be used interchangeably.

Now that those definitions are fresh in our minds, let's look at how pH and pK_{a} are related. The first thing you need to know is that we can use pH and pKa to describe the relationship between **weak acids** in an aqueous solution.

**pH **is a measurement of the [H^{+}] ion concentration in a solution.

You can learn more about pH by reading "**pH Scale**"!

The definition of pK_{a} can sound confusing, especially if you are not familiar with the **acid dissociation constant**, also known as **K _{a}**. So, let's talk about that!

When it comes to weak acids and pH calculation, we need an extra piece of information, the **acid dissociation constant (K _{a}). **K

**Ka** can also be called acid ionization constant, or acidity constant.

The general formula for a monobasic acid can be written as:$\mathrm{HA}\left(\mathrm{aq}\right)\rightleftharpoons {\mathrm{H}}^{+}\left(\mathrm{aq}\right){\mathrm{A}}^{-}\left(\mathrm{aq}\right)$, where:

HA is the weak acid

H+ is the hydrogen ions

A- is the conjugate base.

We can use the following formula for K_{a}:

${\mathrm{K}}_{\mathrm{a}}=\frac{\left[\mathrm{products}\right]}{\left[\mathrm{reactants}\right]}=\frac{\left[{\mathrm{H}}^{+}\right]\left[{\mathrm{A}}^{-}\right]}{\mathrm{HA}}=\frac{{\left[{\mathrm{H}}^{+}\right]}^{2}}{\mathrm{HA}}$

Keep in mind that solids **(s)** and pure liquids** ****(l)** like H_{2}O (l) should not be included when calculating K_{a }because they have constant concentrations. Let's look at an example!

**What would be the equilibrium expression for the following equation? **

Using the formula for K_{a}, the equilibrium expression would be:

${\mathrm{K}}_{\mathrm{a}}=\frac{\left[\mathrm{products}\right]}{\left[\mathrm{reactants}\right]}=\frac{\left[{\mathrm{H}}^{+}\right]\left[{\mathrm{CH}}_{3}{\mathrm{COO}}^{-}\right]}{\left[{\mathrm{CH}}_{3}\mathrm{COOH}\right]}$

For extra practice, try writing the equilibrium expression of ${\mathrm{NH}}_{4}^{+}\left(\mathrm{aq}\right)\rightleftharpoons {\mathrm{H}}^{+}\left(\mathrm{aq}\right)+{\mathrm{NH}}_{3}\left(\mathrm{aq}\right)$ !

Now that we know what K_{a} means, we can define pK_{a. }Don't worry about pK_{a }calculations right now - we will deal with it in a little bit!

**pK _{a} **is referred to as the negative log of K

_{a}can be calculated using the equation:**pK**_{a}= - log_{10}(K_{a})

Buffers are solutions that contain either a weak acid + its conjugate base or a weak base + its conjugate acid, and have the ability to resist changes in pH.

When dealing with buffers, pH and pKa are related through the **Henderson-Hasselbalch** equation, which has the following formula:

$\mathrm{pH}={\mathrm{pK}}_{\mathrm{a}}+\mathrm{log}\frac{[{\mathrm{A}}^{-}]}{\left[\mathrm{HA}\right]}$

The main difference between pH and pK_{a} is that **pK _{a} **is used to show the strength of an acid. On the other hand,

pH | pK _{a} |

$\mathrm{pH}=-{\mathrm{log}}_{10}\left[{\mathrm{H}}^{+}\right]$ | ${\mathrm{pK}}_{\mathrm{a}}=-{\mathrm{log}}_{10}\left[{\mathrm{K}}_{\mathrm{a}}\right]$ |

↑ pH = basic↓ pH = acidic | ↑ pK _{a} = weak acid↓ pK_{a} = strong acid |

depends on [H ^{+}] concentration | depends on [HA], [H ^{+}] and A^{-} |

When we have a strong acid, such as HCl, it will completely dissociate into H^{+} and Cl^{-} ions. So, we can assume that the concentration of [H^{+}] ions will be equal to the concentration of HCl.

$\mathrm{HCl}\to {\mathrm{H}}^{+}+{\mathrm{Cl}}^{-}$

However, calculating the pH of weak acids is not as simple as with strong acids. To calculate the pH of weak acids, we need to use **ICE charts **to determine how many H^{+} ions we will have at equilibrium, and also use **equilibrium expressions (K _{a}). **

$\mathrm{HA}\left(\mathrm{aq}\right)\rightleftharpoons {\mathrm{H}}^{+}\left(\mathrm{aq}\right){\mathrm{A}}^{-}\left(\mathrm{aq}\right)$

**Weak** **acids**** **are those that **partially **ionize in solution.

The easiest way to learn about ICE tables is by looking at an example. So, let's use an ICE chart to find the pH of a 0.1 M solution of acetic acid (The K_{a} value for acetic acid is 1.76 x 10^{-5}).

* Step 1: *First, write down the generic equation for weak acids:$\mathbf{HA}\mathbf{}\mathbf{\left(}\mathbf{aq}\mathbf{\right)}\mathbf{}\mathbf{\rightleftharpoons}\mathbf{}{\mathbf{H}}^{\mathbf{+}}\mathbf{}\mathbf{\left(}\mathbf{aq}\mathbf{\right)}\mathbf{}\mathbf{}{\mathbf{A}}^{\mathbf{-}}\mathbf{}\mathbf{\left(}\mathbf{aq}\mathbf{\right)}$

**Step 2: **Then, create an ICE chart. "I" stands for initial, "C" stands for change, and "E" stands for equilibrium. From the problem, we know that the initial concentration of acetic acid is equal to 0.1 M. So, we need to write that number on the ICE chart. Where? On the "I" row, under HA. Before dissociation, we don't have H^{+} or A^{-} ions. So, write a value of 0 under those ions.

Actually, pure water does have a little bit of H^{+} ions (1 x 10^{-7} M). But, we can ignore it for now since the amount of H^{+} ions that will be produced by the reaction will be way more significant.

** Step 3: **Now, we need to fill out the "C" (change) row. When dissociation occurs, change goes to the right. So, the change in HA will be $-x$, whereas the change in the ions will be $\mathit{+}x$.

* Step 4:* The equilibrium row shows the concentration at equilibrium. "E" can be filled by using the values of "I" and "C". So, HA will have a concentration of 0.1 -

* Step 5: *Now, we have to create an

*x*is equal to the [H^{+}] ion concentration. So, by finding*x*, we will be able to know [H^{+}] and then calculate pH.

${\mathbf{K}}_{\mathbf{a}}=\frac{\left[{\mathrm{H}}^{+}\right]\left[{\mathrm{A}}^{-}\right]}{\mathrm{HA}}=\frac{{\mathbf{x}}^{\mathbf{2}}}{\mathbf{0}\mathbf{.}\mathbf{1}\mathbf{}\mathbf{-}\mathbf{x}}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}$

** Step 6: **Plugin all the known values to the K

${\mathrm{K}}_{\mathrm{a}}=\frac{{\mathrm{x}}^{2}}{0.1{}{-}{\mathrm{x}}}\phantom{\rule{0ex}{0ex}}1.76\times {10}^{-5}=\frac{{\mathrm{x}}^{2}}{0.1}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\mathbf{x}=\sqrt{(1.76\times {10}^{-5}})(0.1)=0.0013\mathrm{M}=\mathbf{}\mathbf{[}{\mathbf{H}}^{\mathbf{+}}\mathbf{]}$

If after doing this step it turns out that x is bigger than 0.05 then you will have to do the whole quadratic equation. After some algebra in this case you would get x^2 +Ka*x - 0.1*Ka = 0. You can just use the normal quadratic formula now to solve for x.

* Step 7:* Use the [H

$\mathrm{pH}=-{\mathrm{log}}_{10}\left[{\mathrm{H}}^{+}\right]\phantom{\rule{0ex}{0ex}}\mathrm{pH}=-{\mathrm{log}}_{10}[0.0013]\phantom{\rule{0ex}{0ex}}\mathrm{pH}=2.9$

Normally, when finding the pH of a weak acid, you will be asked to construct an ICE table. However, for your AP exam (and also to reduce time), there is a little shortcut that you can take to find the [H+] ion concentration of a weak acid that is needed to find its pH.

So, to calculate the [H^{+}] all you need to know is the value for the concentration of the weak acid and the K_{a} value, and plug those values into the following equation:

$\left[{\mathrm{H}}^{+}\right]=\sqrt{{\mathrm{K}}_{\mathrm{a}}\times \mathrm{initial}\mathrm{concentation}\mathrm{of}\mathrm{HA}}$

Then, you can use the [H+] value to calculate pH. Note that this equation will not be given to you in the AP exam, so you should try to memorize it!

To calculate pH and pK_{a}, you should be familiar with the following formulas:

_{ Formulas relating pH and pKa, Isadora Santos - StudySmarter Originals. }

Let's look at a problem!

**Find the pH of a solution containing 1.3x10 ^{-5} M [H^{+}] ion concentration. **

All we have to do is use the first formula above to calculate pH.

$\mathrm{pH}=-{\mathrm{log}}_{10}\left[{\mathrm{H}}^{+}\right]\phantom{\rule{0ex}{0ex}}\mathrm{pH}=-{\mathrm{log}}_{10}\left[1.3\times {10}^{-5}\mathrm{M}\right]\phantom{\rule{0ex}{0ex}}\mathrm{pH}=4.9$

That was pretty straightforward, right? But, let's amp up the difficulty a bit more!

**Find the pH of 0.200 M of benzoic acid. The K _{a} value for C_{6}H_{5}COOH is 6.3 x 10^{-5} mol dm^{-3}.**

${\mathrm{C}}_{6}{\mathrm{H}}_{5}\mathrm{COOH}\stackrel{}{\leftrightharpoons}{\mathrm{H}}^{+}+{\mathrm{C}}_{6}{\mathrm{H}}_{5}{\mathrm{COO}}^{-}$

Although we can make an ICE table to find the [H+] ion concentration of benzoic, let's use the shortcut formula:

$\mathbf{\left[}{\mathbf{H}}^{\mathbf{+}}\mathbf{\right]}\mathbf{}\mathbf{=}\mathbf{}\sqrt{{\mathbf{K}}_{\mathbf{a}}\mathbf{}\mathbf{\times}\mathbf{}\mathbf{initial}\mathbf{}\mathbf{concentation}\mathbf{}\mathbf{of}\mathbf{}\mathbf{HA}}$

So, the value for the hydrogen ion concentration of H^{+} will be:

$\left[{\mathrm{H}}^{+}\right]=\sqrt{(6.3\times {10}^{-5})(0.200)}=0.00355$

Now, we can use the calculated [H^{+}] value to find pH.

$\mathrm{pH}=-{\mathrm{log}}_{10}\left[{\mathrm{H}}^{+}\right]\phantom{\rule{0ex}{0ex}}\mathrm{pH}=-{\mathrm{log}}_{10}[0.00355]\phantom{\rule{0ex}{0ex}}\mathrm{pH}=2.450$

Now, what if you were asked to calculate **pKa from Ka**? All you need to do is use the pK_{a} formula if you know the value for K_{a. }

For example, if you know that the K_{a} value for benzoic acid is 6.5x10^{-5} mol dm^{-3}, you can use it to calculate pK_{a}:

${\mathrm{pK}}_{\mathrm{a}}=-{\mathrm{log}}_{10}\left({\mathrm{K}}_{\mathrm{a}}\right)\phantom{\rule{0ex}{0ex}}{\mathrm{pK}}_{\mathrm{a}}=-{\mathrm{log}}_{10}\left(6.3\times {10}^{-5}\right)\phantom{\rule{0ex}{0ex}}{\mathrm{pK}}_{\mathrm{a}}=4.2$

We can use the pH and concentration of a weak acid to calculate the pK_{a} of the solution. Let's look at an example!

**Calculate the pK _{a} of a 0.010 M solution of a weak acid containing a pH value of 5.3**.

*Step 1: *Use the pH value to find [H^{+}] ion concentration by rearranging the pH formula. By knowing the concentration of [H+], we can also apply it to the concentration of A^{-} since the reaction of weak acids is at equilibrium.

$\left[{\mathrm{H}}^{+}\right]={10}^{-\mathrm{pH}}\phantom{\rule{0ex}{0ex}}\left[{\mathrm{H}}^{+}\right]={10}^{-5.3}=5.0\times {10}^{-6}$

*Step 2: *Make an ICE chart. Remember that "X" is the same as the [H^{+}] ion concentration.

*Step 3: * Write the equilibrium expression using the values in the equilibrium row (E), and then solve for K_{a}.

${\mathbf{K}}_{\mathbf{a}}=\frac{\left[\mathrm{products}\right]}{\left[\mathrm{reactants}\right]}=\frac{\left[{\mathrm{H}}^{+}\right]\left[{\mathrm{A}}^{-}\right]}{\mathrm{HA}}=\frac{{\mathbf{X}}^{\mathbf{2}}}{\mathbf{0}\mathbf{.}\mathbf{010}\mathbf{}\mathbf{-}\mathbf{}\mathbf{X}}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}{\mathrm{K}}_{\mathrm{a}}=\frac{(5.0\times {10}^{-6})(5.0\times {10}^{-6})}{0.010-5.0\times {10}^{-6}}=2.5\times {10}^{-9}\mathrm{mol}{\mathrm{dm}}^{-3}$

*Step 4: *Use the calculated K_{a} to find pK_{a}.

${\mathrm{pK}}_{\mathrm{a}}=-{\mathrm{log}}_{10}{\mathrm{K}}_{\mathrm{a}}=-{\mathrm{log}}_{10}(2.5\times {10}^{-9})\phantom{\rule{0ex}{0ex}}{\mathrm{pK}}_{\mathrm{a}}=8.6$

Another way of measuring the strength of acids is through **percent ionization**. The formula to calculate percent ionization is given as:

$\%\mathrm{ionization}=\frac{\mathrm{Concentration}\mathrm{of}{\mathrm{H}}^{+}\mathrm{ions}\mathrm{in}\mathrm{equilibrium}}{\mathrm{Initial}\mathrm{concentration}\mathrm{of}\mathrm{the}\mathrm{weak}\mathrm{acid}}=\frac{x}{\left[\mathrm{HA}\right]}\times 100$

Remember: *the stronger the acid, the greater the % ionization.* Let's go ahead and apply this formula to an example!

**Find the K _{a }value and the percent ionization of a 0.1 M solution of a weak acid containing a pH of 3. **

1. Use the pH to find [H^{+}].

$\left[{\mathrm{H}}^{+}\right]={10}^{-\mathrm{pH}}\phantom{\rule{0ex}{0ex}}\left[{\mathrm{H}}^{+}\right]={10}^{-3}$

2. Make an ICE table to find the concentrations of HA, H^{+}, and A^{-} in equilibrium.

3. Calculate percent ionization using the value for x ([H+]) and for HA from the ICE table.

$\%ionization=\frac{\left[{\mathrm{H}}^{+}\right]}{\left[\mathrm{HA}\right]}\times 100\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\%\mathrm{ionization}=\frac{[{10}^{-3}\mathrm{M}]}{0.1\mathrm{M}-{10}^{-3}\mathrm{M}}\times 100=\mathbf{1}\mathbf{\%}$

Now, you should have what it takes to find the pH and pK_{a} of weak acids!

**pH**is a measurement of the [H^{+}] ion concentration in a solution.**pK**is referred to as the negative log of K_{a}_{a}.- To calculate the pH and pKa of weak acids, we need to use ICE charts to determine how many H+ ions we will have at equilibrium, and also K
_{a}. - If we know the concentration of H+ ions in equilibrium, and the initial concentration of the weak acid, we can calculate
**percent ionization**.

_{References: }

_{Brown, T. L., Nelson, J. H., Stoltzfus, M., Kemp, K. C., Lufaso, M., & Brown, T. L. (2016). Chemistry: The central science. Harlow, Essex: Pearson Education Limited.}

_{Malone, L. J., & Dolter, T. (2013). Basic concepts of of Chemistry. Hoboken, NJ: John Wiley.}

_{Ryan, L., & Norris, R. (2015). Cambridge International as and A level chemistry. Cambridge: Cambridge University Press.}

_{Salazar, E., Sulzer, C., Yap, S., Hana, N., Batul, K., Chen, A., . . . Pasho, M. (n.d.). Chad's general chemistry Master course. Retrieved May 4, 2022, from https://courses.chadsprep.com/courses/general-chemistry-1-and-2}

**pH** is a measurement of the [H+] ion concentration in a solution. On ther other hand, **pKa** is used to show whether an acid is strong or weak.

In buffers, pH and pKa are related through the **Henderson-Hasselbalch **equation.

**pH** is the negative log (base 10) of [H^{+}]. **pKa** is the negative log (base) of Ka.

More about pH and pKa

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