8.2 - pH and pOH of Strong Acids and Bases

Overview

Strong acids and strong bases are the easy case in Unit 8: they react with water completely, so you can read the ion concentration straight from the label on the bottle. This topic shows how to go from the concentration of a strong acid or base to

[\mathrm{H_3O^+}]

[\mathrm{OH^-}]

, pH, and pOH, using the tools from Topic 8.1.

Strong acids ionize completely

A strong acid gives its proton to water so completely that essentially no acid molecules are left in solution. The six strong acids to memorize:

Strong acid	Name
HCl	hydrochloric acid
HBr	hydrobromic acid
HI	hydroiodic acid
$\mathrm{HNO_3}$	nitric acid
$\mathrm{HClO_4}$	perchloric acid
$\mathrm{H_2SO_4}$	sulfuric acid (first proton only)

Because the reaction goes to completion, it gets a single arrow:

\mathrm{HNO_3(aq) + H_2O(l) \rightarrow H_3O^{+}(aq) + NO_3^{-}(aq)}

Every

\mathrm{HNO_3}

molecule becomes one

\mathrm{H_3O^{+}}

ion and one

\mathrm{NO_3^{-}}

ion (the acid's conjugate base). So for a strong monoprotic acid, the hydronium concentration equals the starting concentration of the acid:

[\mathrm{H_3O^+}] = [\mathrm{HA}]_0 \quad \text{(strong monoprotic acid)}

The major species in the solution, the particles present in large amounts, are

\mathrm{H_3O^{+}}

, the conjugate base anion, and water.

\mathrm{OH^{-}}

is present only in a tiny amount, set by

K_w

About $\mathrm{H_2SO_4}$ : only its first proton is strong. The second one comes from

\mathrm{HSO_4^{-}}

, which is a weak acid, so a sulfuric acid solution has somewhat more

\mathrm{H_3O^{+}}

than its concentration, but less than double. AP questions that ask for an exact pH usually stick to the monoprotic strong acids.

Once you have

[\mathrm{H_3O^+}]

, the rest follows from 8.1: pH =

-\log[\mathrm{H_3O^+}]

, then pOH = 14.00

-

pH, and

[\mathrm{OH^-}] = K_w/[\mathrm{H_3O^+}]

. Strong acids more concentrated than 1 M even have negative pH values: 2.0 M HCl has pH =

-\log(2.0) = -0.30

Example 1: A strong acid solution

Find

[\mathrm{H_3O^+}]

, pH, pOH, and

[\mathrm{OH^-}]

in 0.0250 M HBr at 25 °C. What are the major species?

Step 1: HBr is a strong acid, so it ionizes completely.

\mathrm{HBr(aq) + H_2O(l) \rightarrow H_3O^{+}(aq) + Br^{-}(aq)}

[\mathrm{H_3O^+}] = 0.0250\ \mathrm{M}

Step 2: Take the log. Three significant figures give three decimal places.

\mathrm{pH} = -\log(0.0250) = 1.602

Step 3: Find pOH and

[\mathrm{OH^-}]

K_w

has two significant figures, so results that come from it get two.

\mathrm{pOH} = 14.00 - 1.602 = 12.40 \qquad [\mathrm{OH^-}] = \frac{1.0 \times 10^{-14}}{0.0250} = 4.0 \times 10^{-13}\ \mathrm{M}

Answer:

[\mathrm{H_3O^+}] = 0.0250

M, pH = 1.602, pOH = 12.40, and

[\mathrm{OH^-}] = 4.0 \times 10^{-13}

M. The major species are

\mathrm{H_3O^{+}}

\mathrm{Br^{-}}

, and

\mathrm{H_2O}

. There are essentially no HBr molecules left.

Notice what dilution does. Since

[\mathrm{H_3O^+}]

equals the acid concentration, diluting a strong acid tenfold cuts

[\mathrm{H_3O^+}]

by a factor of 10 and raises the pH by exactly 1 unit.

Strong bases dissociate completely

The strong bases are the group 1 hydroxides (LiOH, NaOH, KOH, RbOH, CsOH) and the group 2 hydroxides (the common ones are

\mathrm{Ca(OH)_2}

\mathrm{Sr(OH)_2}

, and

\mathrm{Ba(OH)_2}

). They're ionic compounds, so the ions already exist in the solid. Dissolving them just separates the ions, a process called dissociation:

\mathrm{NaOH(s) \rightarrow Na^{+}(aq) + OH^{-}(aq)}

\mathrm{Ba(OH)_2(s) \rightarrow Ba^{2+}(aq) + 2OH^{-}(aq)}

The difference that matters is the 2 in front of

\mathrm{OH^{-}}

. Each formula unit of a group 2 hydroxide releases two hydroxide ions:

8.2 - pH and pOH of Strong Acids and Bases

Both boxes started with 4 formula units (water isn't shown). The NaOH box has 4

\mathrm{OH^{-}}

, but the

\mathrm{Ba(OH)_2}

box has 8. So:

[\mathrm{OH^-}] = [\mathrm{MOH}]_0 \ \ \text{(group 1)} \qquad [\mathrm{OH^-}] = 2[\mathrm{M(OH)_2}]_0 \ \ \text{(group 2)}

Then find pOH =

-\log[\mathrm{OH^-}]

and pH = 14.00

-

pOH.

\mathrm{Ca(OH)_2}

doesn't dissolve much, but whatever does dissolve is completely dissociated, so it still counts as a strong base.

Example 2: Barium hydroxide

Find

[\mathrm{OH^-}]

, pOH, pH, and

[\mathrm{H_3O^+}]

in 0.00750 M

\mathrm{Ba(OH)_2}

at 25 °C.

Step 1: Each

\mathrm{Ba(OH)_2}

gives 2

\mathrm{OH^{-}}

[\mathrm{OH^-}] = 2(0.00750\ \mathrm{M}) = 0.0150\ \mathrm{M}

Step 2: Take the log, then use pH + pOH = 14.00.

\mathrm{pOH} = -\log(0.0150) = 1.824 \qquad \mathrm{pH} = 14.00 - 1.824 = 12.18

Step 3: Find

[\mathrm{H_3O^+}]

from

K_w

[\mathrm{H_3O^+}] = \frac{1.0 \times 10^{-14}}{0.0150} = 6.7 \times 10^{-13}\ \mathrm{M}

Answer:

[\mathrm{OH^-}] = 0.0150

M, pOH = 1.824, pH = 12.18, and

[\mathrm{H_3O^+}] = 6.7 \times 10^{-13}

M. Forgetting the 2 would give pH 11.88, which is wrong.

Example 3: Starting from a mass (AP-style)

A student dissolves 0.350 g of KOH (

\mathcal{M}

= 56.11 g/mol) in enough water to make 250.0 mL of solution at 25 °C. (a) Calculate the pH. (b) What concentration of

\mathrm{Ba(OH)_2}

would have the same pH? (c) The student dilutes 10.00 mL of the KOH solution to 100.0 mL. Predict the new pH and justify it without a full calculation.

(a) Convert grams to moles, then divide by liters.

n = \frac{0.350\ \mathrm{g}}{56.11\ \mathrm{g/mol}} = 6.24 \times 10^{-3}\ \mathrm{mol} \qquad [\mathrm{OH^-}] = \frac{6.24 \times 10^{-3}\ \mathrm{mol}}{0.2500\ \mathrm{L}} = 0.0250\ \mathrm{M}

\mathrm{pOH} = -\log(0.0250) = 1.602 \qquad \mathrm{pH} = 14.00 - 1.602 = 12.40

(b) Same pH means the same

[\mathrm{OH^-}]

, 0.0250 M. Each

\mathrm{Ba(OH)_2}

gives two

\mathrm{OH^{-}}

, so it takes half as much:

[\mathrm{Ba(OH)_2}] = 0.0250/2 = 0.0125

(c) Going from 10.00 mL to 100.0 mL is a tenfold dilution, so

[\mathrm{OH^-}]

drops by a factor of 10 to 0.00250 M. A factor of 10 is one log unit, so pOH rises by 1 and pH falls by 1, to 11.40.

Summary of strong acid and base solutions

Solution (concentration $C$ )	Major species	Ion concentration
Strong monoprotic acid, like $\mathrm{HNO_3}$	$\mathrm{H_3O^{+}}$ , $\mathrm{NO_3^{-}}$ , $\mathrm{H_2O}$	$[\mathrm{H_3O^+}]$ = $C$
Group 1 hydroxide, like NaOH	$\mathrm{Na^{+}}$ , $\mathrm{OH^{-}}$ , $\mathrm{H_2O}$	$[\mathrm{OH^-}]$ = $C$
Group 2 hydroxide, like $\mathrm{Ba(OH)_2}$	$\mathrm{Ba^{2+}}$ , $\mathrm{OH^{-}}$ , $\mathrm{H_2O}$	$[\mathrm{OH^-}]$ = $2C$

Common mistakes

Forgetting the 2 for group 2 hydroxides. $\mathrm{Ba(OH)_2}$ and $\mathrm{Sr(OH)_2}$ give two $\mathrm{OH^{-}}$ per formula unit.

Reporting pOH as pH. For a base you usually find pOH first. Finish with pH = 14.00 $-$ pOH.

Calling HF a strong acid. HF is weak, even though HCl, HBr, and HI are strong.

Doubling $[\mathrm{H_3O^+}]$ for $\mathrm{H_2SO_4}$ . Only the first proton ionizes completely.

Making an acid solution basic. $1.0 \times 10^{-8}$ M HCl doesn't have pH 8. That's less acid than water's own $1.0 \times 10^{-7}$ M of $\mathrm{H_3O^{+}}$ , so the pH is just under 7 (about 6.98). Adding acid can never make a solution basic.

Confusing strong with concentrated. "Strong" means it ionizes completely. A very dilute HCl solution is still a strong acid.

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