Topic 8.1 Notes – Introduction to Acids and Bases

1. What pH and pOH Actually Measure

When we talk about acidity, we are measuring the concentration of hydronium ions, $\ce{H3O+}$, in solution.

$${pH = -\log[\ce{H3O+}]}$$

For bases, we measure hydroxide ions, $\ce{OH-}$:

$${pOH = -\log[\ce{OH-}]}$$

A few log facts that save time on quizzes:

• If $[\ce{H3O+}] = 1.0 \times 10^{-4}$, then pH = 4.00
• A lower pH means a higher $[\ce{H3O+}]$
• Each 1-unit drop in pH means 10× more hydronium
  (pH 3 is 10× more acidic than pH 4)

H⁺ vs. H₃O⁺

In water, a free proton does not exist alone. It attaches to water:

$$\ce{H+ + H2O -> H3O+}$$

On the AP exam, $\ce{H+}$ and $\ce{H3O+}$ are treated as the same for calculations. Just remember that chemically, $\ce{H3O+}$ is more accurate.

2. Water Autoionization and $K_w$

Water is not completely neutral in the sense of “no ions.” It slightly reacts with itself:

$$\ce{2H2O(l) <=> H3O+(aq) + OH-(aq)}$$

This is called autoionization.

The equilibrium constant for this process is:

$${K_w = [\ce{H3O+}][\ce{OH-}]}$$

At 25°C:

$${K_w = 1.0 \times 10^{-14}}$$

This number is huge conceptually. It means:

• In any aqueous solution,
  $$[\ce{H3O+}][\ce{OH-}] = 1.0 \times 10^{-14} \quad (25^\circ\text{C})$$
• If you know one concentration, you can always find the other.

Example:

If $[\ce{H3O+}] = 2.0 \times 10^{-5}$,

$$[\ce{OH-}] = \frac{1.0 \times 10^{-14}}{2.0 \times 10^{-5}} = 5.0 \times 10^{-10}$$

That reciprocal relationship shows up constantly in MCQs.

3. Neutral Solutions and the pH + pOH Relationship

In pure water, every hydronium formed creates one hydroxide. So:

$$[\ce{H3O+}] = [\ce{OH-}]$$

Using $K_w$:

$$[\ce{H3O+}]^{2} = 1.0 \times 10^{-14}$$

$$[\ce{H3O+}] = 1.0 \times 10^{-7} \text{ M}$$

So at 25°C:

• pH = 7.00
• pOH = 7.00
• The solution is neutral

Taking the negative log of $K_w$:

$${pK_w = pH + pOH}$$

At 25°C:

$${pH + pOH = 14.00}$$

This applies to any aqueous solution at 25°C, not just pure water.

All of those relationships connect pH, pOH, $[\ce{H3O+}]$, and $[\ce{OH-}]$ in a tight loop:

pH, pOH, $[\ce{H3O+}]$, and $[\ce{OH-}]$ conversion relationships at 25°C

Focus on the top connection: $\text{pH} + \text{pOH} = 14.00$ at 25°C. When one goes up, the other must go down.

4. Temperature Dependence of $K_w$

Here’s where students slip up.

$K_w$ changes with temperature.

• If temperature increases → $K_w$ increases
• If temperature decreases → $K_w$ decreases

Neutral means:

$$[\ce{H3O+}] = [\ce{OH-}]$$

It does not mean pH = 7.

At temperatures above 25°C:

• $K_w > 1.0 \times 10^{-14}$
• Neutral pH is less than 7

At temperatures below 25°C:

• $K_w < 1.0 \times 10^{-14}$
• Neutral pH is greater than 7

On most tests, if temperature is not mentioned, assume 25°C. But if they give you a different temperature and a different $K_w$, do not automatically use 14.

5. Solving pH and pOH Problems

You should be able to move between four forms:

$$[\ce{H3O+}] \leftrightarrow pH \leftrightarrow pOH \leftrightarrow [\ce{OH-}]$$

Given $[\ce{H3O+}]$

$$pH = -\log[\ce{H3O+}]$$

$$pOH = 14 - pH \quad (25^\circ\text{C})$$

Given $[\ce{OH-}]$

$$pOH = -\log[\ce{OH-}]$$

$$pH = 14 - pOH$$

Given pH

$$[\ce{H3O+}] = 10^{-pH}$$

Given pOH

$$[\ce{OH-}] = 10^{-pOH}$$

Be careful with significant figures. The number of decimal places in pH equals the significant figures in concentration. That detail shows up on FRQs.