**Acids and bases** are key components of many chemical reactions and systems. To know chemistry, one must know **acid-base chemistry** — and the two __always__ go together.

Any chemical substance can either be **acidic** (strongly or weakly), **neutral**, or **basic** (strongly or weakly). Here are a few examples of acids, bases and neutral compounds along a continuum, from strong acid to strong base.

We will encounter other, more subtle definitions of acids and bases later, but for now, a good working definition that is useful most of the time is:

Acidsare H^{+}(proton) donors

Basesare OH^{-}(hydroxyl ion) donors

Acids are substances that either directly or indirectly raise the concentration of H^{+} ions (protons) in solution. Bases either directly or indirectly raise the concentration of OH^{-} (hydroxyl ions) in solution.

You will want to *memorize* the strongest acids and bases (table below).

Strong acids and bases are those which are completely soluble in water.

HCl, HBr, HI– thehalogenicacids

H– sulfuric acid_{2}SO_{4}

HNO– nitric acid_{3}

HClO– perchloric acid_{4}

KOH– potassium hydroxide

NaOH– sodium hydroxide

Ba(OH)– barium hydroxide_{2}

Ca(OH)– calcium hydroxide_{2}

X
### "strong"

In the sense of acids and bases, "strong" means that the compound completely dissociates in water. Weak acids and bases only dissociate partly, or are *insoluble*.

HCl, hydrochloric acid*, is a strong acid—one of the strongest. It dissociates completely into a chloride ion and a free proton (from here on, I'll call a +1 hydrogen ion a **proton**, because that's what it is).

It's the free proton that makes a strong acid different from a neutral compound. If one mole of HCl is dissolved into one liter of water, the concentration of H+ will be 1M because HCl dissociates completely (it is a **strong electrolyte **- that's the reason for the single arrow in the** dissociation equation**). The same is true of the other acids on the strong acid list. We say that HCl "donates" its proton to the solution—very willingly in this case.

**HCl is a gas at room temperature and pressure.*

*Image source: http://www.chemicallabels.com*

*Image source: http://www.chemicallabels.com*

Sodium hydroxide, NaOH, is one of the most commonly used strong bases. It, too, dissolves completely in water:

It forms a sodium cation and a hydroxyl anion, OH-. NaOH is said to "donate" a hydroxyl ion to the solution.

Sodium hydroxide is a common component of drain cleaners, and is also known by the common name, "lye". It is a white solid at room temperature and pressure. Base compounds are also called **alkali** compounds from time to time.

Water is the prototypical neutral compound. Pure water is what we define neutral to be. It is neither acidic nor basic. A closer look later will show us that it's actually both a weak acid and a weak base. We understand that, unlike the H-atom in HCl, the H-atoms in H_{2}O are bound covalently,

and don't come off easily. Water is not an ionic compound, but it still dissociates. At any given time in a beaker of water, some molecules may be found undergoing the dissociation, known as the **auto-ionization** of water:

A key feature of acids and bases is that they can neutralize one-another. Consider the reaction between the strong acid HCl and the strong base NaOH:

This is typical of a neutralization reaction. The products of neutralization are a salt consisting of the liberated cation of the base and the anion of the acid, and water. Because simple salts don't contain ionizable protons or hydroxyl groups, they are neutral in solution, and of course water is neutral. A solution of NaCl in water, for example, is neutral.

We noted above that at any given time, some water molecules are dissociated into OH- and H+ according to the equation

Often, that equation is written as

where **H _{3}O^{+}** is called the

We can write the equilibrium constant expression for this reaction like this:

Now in a neutral solution **[H ^{+}] = [OH^{-}]**, so

**[H ^{+}][OH^{-}] = [H^{+}][H^{+}] = [H^{+}]^{2} = 1.0 x 10^{-14}**,

thus in a neutral solution, **[H ^{+}] = 1.0 x 10^{-7}**.

Now let's do something that may seem odd, ... and I'll explain later: Let's take the negative of the base-10 log of both sides of that last equation:

**-log _{10} [H^{+}]^{2} = -2 log_{10} [H^{+}] = -log_{10} (1.0 x 10^{-14})**,

which gives

**[H ^{+}] = 7**.

We call this number, the negative of the base-ten log of the proton concentration, the **pH** of the solution.** The pH of pure water is 7**, and we refer to **pH = 7** as the **neutral pH**.

In chemistry, the letter p stands for "take the negative base-ten log," -log_{10}, for example:

pH = -log[H^{+}]

pOH = -log[OH^{-}]

pK_{w} = -log[K_{w}]

Although we could use either pH or pOH to characterize the acidity or basicity of a solution, we use pH by convention. The reason we use the pH scale — a logarithmic scale — at all is that the range of possible proton concentrations is very broad, from nearly 10^{-14} M up to 10 M or more. On the pH scale a very acidic solution might have a pH of 1 while a very basic solution might have a pH of 14. These are manageable numbers, if we can just remember where they came from. At any rate, the pH scale is entrenched in how we do chemistry today.

Also note that a pH 4 solution has 10 times the proton concentration as a pH 5 solution. That's how a logarithmic scale works.

pH < 7 means acidic.

pH > 7 means basic.

pH = 7 means neutral.

A little algebra gives us a handy relation: Notice that because in aqueous solution, **[H ^{+}][OH^{-}] = 1.0 x 10^{-14}**, we can always calculate

**-log ( [H ^{+}][OH^{-}]) = -log(1.0 x 10^{-14})**

Using one of the laws of logs, log(ab) = log(a) + log(b), we get

**-log[H ^{+}] - log[OH^{-}] = -log(1.0 x 10^{-14})**

Now converting the -logs to "p" notation we obtain a very useful relationship between pH and pOH, one that you should memorize:

**pH + pOH = 14**

**Solution****HBr → H+ + Br-**. Because this is a strong acid, we assume that it dissociates completely, so the concentration of H+ in solution will be [H+] = 0.001 M.

$$ \begin{align} pH &= -log(0.001) = 3 \\ pOH &= 14 - pH = 11 \end{align}$$

**Solution****KOH → K+ + OH-**. Because this is a strong base, we assume that it dissociates completely, so that the concentration of OH- in solution will be [OH-] = 0.00015.

$$ \begin{align} pOH &= -log(0.00015) = 3.82 \\ pH &= 14 - pOH = 10.2 \end{align}$$

Strong acids and bases have a lot of street cred. They eat through things, and that's cool, but weak acids and bases are the business end of chemistry, especially in biological systems. Weak acids and bases are a little more difficult to deal with because they don't dissociate completely in solution. That means we have to be mindful of equilibria in which acids and bases aren't completely dissolved.

The proton concentration in a 1.0 M solution of the weak acid acetic acid **(CH _{3}COOH)**, for example, will be much lower than 1.0 M (what we would expect from complete dissociation to CH

Let's do an example with **acetic acid**: Calculate the pH of a 0.1 M solution of acetic acid.

Acetic acid (left) is a weak organic acid known as a carboxylic acid. There are many such organic acids, differing by the identity of what is the CH_{3} group in the figure. If that is an H atom, for example, the acid is **formic acid**, the acid that gives red ant bites their sting.

The carboxyl group, **-COOH**, holds the acidic proton, the one that ionizes relatively easily, though not nearly as easily as the proton of a strong acid.

The dissociation equilibrium is

**CH _{3}COOH ⇌ CH_{3}COO^{-} + H^{+}**

(note the double arrow this time), with:

Here the equilibrium constant is written as K_{a} (short for "K-acid"). These and the K_{b}'s of weak bases have been measured for almost any weak acid or base you will encounter. You can download a table of them here.

Now in a 0.1 M solution of acetic acid (often abbreviated **CH _{3}COOH**), very little

If x moles of **CH _{3}COOH** dissociate, then there will be 0.1-x moles of undissociated

Multiplying both sides of this equation by the denominator on the left and gathering terms on one side, we get this quadratic equation:

**x ^{2} + 1.614 x 10^{-5} x - 1.614 x 10^{-5} = 0**.

This can be solved by completing the square or by using a quadratic equation (Ax^{2} + Bx + C = 0) program on a calculator or computer, with A = 1, B =1.614 x 10^{-5}, C = -1.614 x 10^{-5}. We get x = 0.00252M.

Now pH = -log[H^{+}], so pH = -log[0.00252] = 2.6. For comparison, the pH of a 0.1 M solution of a strong acid would be 1.

It's actually not necessary to solve the quadratic equation every time for weak acids. It turns out that the error in making a certain approximation is usually so small that we can tolerate it, and the approximation makes life with weak acids and bases easier. It's outlined below.

In example 3 above, we needed to solve the equation

which is a quadratic equation. While this isn't too difficult, especially if you've programmed your calculator to do it, there is a convenient approximation that will make life easier. Look at that equation again:

Now x is the concentration of protons, which, for a weak acid, ought to be very small.

We make the approximation that x is insignificant compared to 0.1 and just cross it out:

Now we just need to multiply both sides by 0.1 and take a square root. In this case the solutions are:

Exact pH = 2.598

with approximation pH = 2.896,

an error of about 10%.

As the dissociation constant becomes smaller (for weaker acids), this approximation gets better.

One of the most important weak bases is ammonia. It isn't a prototypical base because it doesn't *directly* supply a solution with OH^{-} ions. Here is the aqueous equilibrium reaction:

We need to pause here to refine our definitions of acids and bases, because our previous one won't work for ammonia. This is the **Brönsted-Lowry** definition of acids and bases:

Acidsare H^{+}(proton) donors

Basesare proton acceptors

This definition covers ammonia. Take another look at the dissociation equation. Ammonia "accepts" a proton from water (note that water acts as an acid in this case), leading indirectly to an increase in hydroxyl ion concentration. The K_{b} for ammonia is K_{b} = 1.778 x 10^{-5}. The K_{b} expression is

Just like K_{a} values, K_{b}s are tabulated for many weak bases. Wikipedia is a pretty good source for chemical properties of all kinds, including acid-base properties.

_{3}.

The dissociation reaction is **NH _{3} + H_{2}O ⇌ NH_{4}^{+} + OH^{-}**. Because this is a weak base, we know it doesn't dissociate completely, so we have to use the equilibrium constant expression (above). If we let x be the amount of NH

Here we have employed our approximation to get

**x = [OH ^{-}] = 0.001333 M**

Now we can calculate the pOH:

**pOH = -log(0.001333) = 2.88**

... and finally the pH of the basic solution:

**pH = 14 - pOH = 11.12**

Solution here

1. |
Calculate the pH of a 1.0 × 10 ## SolutionHCl is a strong acid, HCl → H $$ \begin{align} [H^+] &= 1.0 \times 10^{-3} \; M \\ pH &= -log[ H^+ ] \\ &= -log [10^{-3}] = \bf 3 \end{align}$$ pH 3 is a low pH (< 7), therefore this is an acidic solution. |

2. |
Calculate the pH of a 0.9 M solution of hydrobromic acid (HBr) ## SolutionHBr is a strong acid, HBr → H $$ \begin{align} [H^+] &= 9.0 \times 10^{-1} \; M \\ pH &= -log[ H^+ ] \\ &= -log [0.9] = \bf 0.046 \end{align}$$ pH 0.046 is a very low pH (< 7 and near the lowest pH possible), therefore this is a very acidic solution. |

3. |
Calculate the pH of a 2.234 × 10 ## SolutionKOH is a strong base, KOH → K $$ \begin{align} [OH^-] &= 2.234 \times 10^{-6} \; M \\ pOH &= -log[ OH^- ] \\ &= 5.65 \\ pH &= 14 - pOH = \bf 8.35 \end{align}$$ pH 8.35 is a high pH (> 7), therefore this is a basic solution. |

4. |
Calculate the pH of 735 liters of a solution containing 0.34 moles of nitric acid (HNO ## SolutionFirst we have to calculate the concentration of the strong acid HNO $$\frac{0.34 \; mol \; HNO_3}{735 \, L} = 4.626 \times 10^{-4} \; M$$ Now HNO $$ \begin{align} [H^+] &= 4.626 \times 10^{-4} \; M \\ pH &= -log[ H^+ ] \\ &= -log [4.626 \times 10^{-4}] \\ &= \bf 3.33 \end{align}$$ Despite the large dilution factor (a small number of moles into 735 Liters), this is still quite an acidic solution, pH < 7. |

5. |
Calculate the pH of a 2.0 liter solution containing 0.005 g of HCl. ## SolutionFirst we need to calculate the concentration of the HCl. We have to convert 0.005g to moles, then divide by 2.0 Liters. $$ \begin{align} [HCl] &= \frac{0.005 \, g \; HCl \left( \frac{1 \; mol \; HCl}{37.45 \, g \; HCl}\right)}{2.0 \, L \; solution} \\ \\ &= 6.68 \times 10^{-5}\,M \end{align}$$ Now use that concentration, together with the fact that you kow HCl is a strong acid, which will dissociate completely, to find the pH: $$ \begin{align} [H^+] &= 6.68 \times 10^{-5}\,M \\ pH &= -log[ H^+ ] \\ &= -log [6.68 \times 10^{-5}] \\ &= \bf 4.18 \end{align}$$ |

6. |
Calculate the pH and pOH of a solution with a volume of 5.4 liters that contains 15 g of hydrochloric acid (HCl) and 25 g of nitric acid (HNO ## SolutionThere are two strong acids in this solution. We need to calculate the total number of moles of protons, then divide that by 5.4 L to get [H $$ \begin{align} 15 \, g \; HCl \left( \frac{1 \; mol \; HCl}{36.46 \, g \; HCl} \right) &= 0.411 \, mol \\ \\ 25 \, g \; HNO_3 \left( \frac{1 \; mol \; HNO_3}{63 \, g \; HNO_3} \right) &= 0.397 \, mol \end{align}$$ So the total number of moles is 0.411 + 0.397 mol = 0.808 mol. Now, $$[H^+] = \frac{0.808 \, mol \; H^+}{5.4 \, L} = 0.1495 \, M$$ So the pH is $$ \begin{align} pH &= -log[ H^+ ] \\ &= -log [0.1495 \, M] \\ &= \bf 0.82 \; \; \longleftarrow \; \text{ quite acidic} \\ pOH &= 14 - pH = 14 - 0.82 = 13.18 \end{align}$$ |

7. |
A swimming pool has a volume of 1 million liters. How many grams of HCl would you need to add to the pool to bring the pH from 7 down to 4? Assume that the volume of the HCl is negligible. ## SolutionAt pH 7, there will be no HCl present, assuming we're working with pure H $$ \begin{align} [H^+] &= \frac{x \; mol \; H^+}{10^6 \, L} \; \rightarrow \\ \\ x &= (1 \times 10^{-4})(1 \times 10^6) \\ &= 100 \; mol \; H^+ \end{align}$$ Now for every 1 mol of H $$ \begin{align} 10^2 \; mol \; H^+ &\left( \frac{1 \, mol \; HCl}{1 \, mol \; H^+} \right) \left( \frac{36.46 \, g \; HCl}{1 \; mol \; HCl} \right) \\ \\ &= 3646 \, g \; HCl \\ &= \text{ 3.646 Kg HCl} \end{align}$$ |

8. |
Calculate the pH and pOH of a solution that was made by adding 400 ml of water to 350 ml of a 5.0 × 10 ## SolutionFirs we need the number of moles of NaOH in 350 ml (0.350 L) of a 1.74 × 10 $$0.350 \, L \left( \frac{5.0 \times 10^{-3} \; mol \; NaOH}{1 \; L}\right) \\ \\ = 1.75 \times 10^{-3} \text {moles NaOH}$$ Now the new concentration of NaOH, after dilution is: $$\frac{1.75 \times 10^{-3} \; mol\; NaOH}{(0.350 + 0.400) \, L} = 2.33 \times 10^{-2} \, M$$ So the pOH and pH are: $$ \begin{align} pOH &= -log(2.33 \times 10^{-2}) = 1.63 \\ pH &= 14 - pOH = 12.4 \end{align}$$ |

9. |
The active ingredient of aspirin is acetyl salicylic acid (C - Calculate the pH of a solution made by dissolving 500 mg of ASA in water, and diluting it to 50 ml.
- Repeat the calculation, this time dissolving the ASA in 1000 ml of water.
- Is the pH higher or lower in the more dilute solution? In which solution is the fraction of ionized ASA higher?
## SolutionWe can abbreviate acetylsalicilic acid as HA, where A is the anion and H is the proton that comes off. The dissociation reaction is then HA ⇌ H |

10. |
Papaverine hydrochloride (pap-HCl) is a salt of a weak base (papaverine) and a strong acid (HCl). It is a drug used as a muscle relaxant. Pap-HCl is a weak acid overall. At 25˚C, a 205 mM solution of pap-Hl has a pH of 3.31. Compute the K |

11. |
A 0.040 M solution of a weak acid (call it HA) has a pH of 4.70. Calculate the K ## SolutionSolution here |

**xaktly.com** by Dr. Jeff Cruzan is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License. © 2012, Jeff Cruzan. All text and images on this website not specifically attributed to another source were created by me and I reserve all rights as to their use. Any opinions expressed on this website are entirely mine, and do not necessarily reflect the views of any of my employers. Please feel free to send any questions or comments to jeff.cruzan@verizon.net.