# 15.4: Strong Acids and Strong Bases

Skills to Develop

• Give the names and formulas of some strong acids and bases.
• Explain the pH scale, and convert pH and [HSUP>+].
• Evaluate solution pH and pOH of strong acids or bases.

Acids and bases that are completely ionized when dissolved in water are called strong acids and strong bases There are only a few strong acids and bases, and everyone should know their names and properties. These acids are often used in industry and everyday life. The concentrations of acids and bases are often expressed in terms of pH, and as an educated person, you should have the skill to convert concentrations into pH and pOH. The pH is an indication of the hydrogen ion concentration, $$\ce{[H+]}$$.

The general reaction of an acid with water is given by the general expression:

$\ce{HA}_{(aq)}+\ce{H2O}_{(l)}⇌\ce{H3O+}_{(aq)}+\ce{A-}_{(aq)} \label{gen ion}$

Water is the base that reacts with the acid HA, A is the conjugate base of the acid HA, and the hydronium ion is the conjugate acid of water. A strong acid yields 100% (or very nearly so) of $$\ce{H3O+}$$ and A when the acid ionizes in water; Figure 14.4.1 lists several strong acids and strong bases.

Figure $$\PageIndex{1}$$: Some of the common strong acids and bases are listed here.

#### Strong Acids

Strong acids are acids that are completely or nearly 100% ionized in their solutions; Table $$\PageIndex{1}$$ includes some common strong acids. Hence, the ionization in Equation $$\ref{gen ion}$$ for a strong acid HA can be represented with a single arrow:

$\ce{HA}_{(aq)}+\ce{H2O}_{(l)} \rightarrow \ce{H3O+}_{(aq)}+\ce{A-}_{(aq)}$

For a strong acid, $$\ce{[H+]}$$ = $$\ce{[A-]}$$ = concentration of acid if x is much greater than $$1 \times 10^{-7}$$. For a very dilute strong acid solution with concentration less than $$1 \times ^{-7}$$, the pH is dominated by the autoionization of water

$\ce{H2O \rightleftharpoons H+ + OH-}, \hspace{10px} \mathrm{K_w} = 1 \times 10^{-14} \text{at 298 K}$

Example $$\PageIndex{1}$$

Calculate the pH of a solution with $$1.2345 \times 10^{-4}\; M \ce{HCl}$$

SOLUTION
The solution of a strong acid is completely ionized. Thus, $$\ce{[H+]} = 1.234 \times 10^{-4}$$.

$\ce{pH} = -\log(1.234 \times 10^{-4}) = 3.909$

Exercise $$\PageIndex{1}$$

What is the pH for a solution containing 1.234 M $$\ce{[HCl]}$$?

pH = 0.0913

Example $$\PageIndex{2}$$

Calculate the pH of a stock $$\ce{HCl}$$ solution that is 32% by mass $$\ce{HCl}$$.

SOLUTION
The density of such a solution is needed before we can calculate the pH. Since the density is not on the label, we need to find it from the Material Safety Data Sheet, which gives the specific gravity of 1.150. Thus, the amount of acid in 1.0 L is 1150 g.

\begin{align} \textrm{The amount of HCl} &= 1000\times1.150\times0.32\\ &= \mathrm{368\: g\: \left(\dfrac{1\: mol}{36.5\: g} \leftarrow molar\: mass\: of\: HCl\right)}\\ &= \mathrm{10.08\: M}\\ &= \ce{[H+]} \end{align}

$\ce{pH} = -\log(10.08) = -1.003$

DISCUSSION
Yes, pH have negative values if $$\ce{[H+]} > 1.0$$

Exercise $$\PageIndex{2}$$

Check out the information on nitric acid, and calculate the pH of a stock nitric acid solution.

Example $$\PageIndex{3}$$

Calculate the pH of a solution containing $$1.00 \times 10^{-7}\; M$$ of $$\ce{HCl}$$ .

SOLUTION
$$\ce{[H+]} = 1.0 \times 10^{-7}\; M$$ from the strong acid, and if x is the amount from the ionization of water, then we have the equilibrium due to the autoionization of water. We can model this with an ICE table.

$H_2O \rightleftharpoons H^+_{(aq)} + OH^-_{(aq)}$

ICE Table $$H_2O_{(l)}$$ $$H^+_{(aq)}$$ $$OH^-_{(aq)}$$
Initial - $$1\times 10^{-7}$$ 0
Change - +x +x
Equilibrium - $$1\times 10^{-7} + x$$ x

Recall that $$K_{w} = \ce{[H+] [OH- ]} = 1 \times 10^{-14}$$, due to the ionization equilibrium of water in the solution:

$\{(1.00 \times 10^{-7} +x) x = 1 \times 10^{-14}$

$x^2 + 1.00 \times 10^{-7}x - 1.00 \times 10^{-14} = 0$

Solving this equation for $$x$$ from the quadratic equation results in

\begin{align} x &= \dfrac{-1.00 \times 10^{-7} \pm \sqrt{1.00 \times 10^{-14} + (4)(1) (1.00 \times 10^{-14})}}{2}\\ &= 0.61 \times 10^{-7} \end{align}

only the additive root is physical (positive concentration); therefore

$\ce{[H+]} = (1.00 + 0.61) \times 10^{-7}\; M$

$\ce{pH} = -\log(1.61 \times 10^{-7}) = 6.79$

DISCUSSION
If you require only 1 significant figure, the pH is about 7.

#### Strong Bases

Strong bases are completely ionized in solution; Table 16.4.1 includes some common strong bases. For example, $$\ce{KOH}$$ dissolves in water in the reaction

$\ce{KOH \rightarrow K+ + OH-}$

Relative to the number of strong acids, there are fewer number of strong bases and most are alkali hydroxides. Calcium oxide is considered a strong base, because it is completely, almost completely, ionized. However, the solubility of calcium hydroxide is very low. When $$\ce{Ca(OH)2}$$ dissolves in water, the ionization reaction is as follows:

$\ce{Ca(OH)2 \rightarrow Ca^2+ + 2 OH-}$

The concentration of $$\ce{OH-}$$ is twice the concentration of $$\ce{Ca^2+}$$,

$\mathrm{[OH^-] = 2 [Ca^{2+}]}$

Example $$\PageIndex{4}$$

Calculate the pOH of a solution containing $$1.2345 \times 10^{-4}\; M \; \ce{Ca(OH)2}$$.

SOLUTION
Based on complete ionization of

$\ce{Ca(OH)2 \rightarrow Ca^{+2} + 2OH-}$

\begin{align} \ce{[OH-]} &= 2 \times 1.234 \times 10^{-4} = 2.468 \times 10^{-4}\; M \\ \ce{pOH} &= -\log( 2.468 \times 10^{-4})\\ &= 3.61 \end{align}

Exercise $$\PageIndex{4}$$

The molar solubility of calcium hydroxide is 0.013 M $$\ce{Ca(OH)2}$$. Calculate the pOH.

pOH = 1.58

### Questions

1. What is the pH of a solution containing 0.01 M $$\ce{HNO3}$$?
2. What is the pH of a solution containing 0.0220 M $$\ce{Ba(OH)2}$$? Give 3 significant figures.
3. Exactly 1.00 L solution was made by dissolving 0.80 g of $$\ce{NaOH}$$ in water. What is $$\ce{[H+]}$$? (Atomic mass: $$\ce{Na}$$, 23.0; $$\ce{O}$$, 16.0; $$\ce{H}$$, 1.0)
4. What is the pH for a solution which is 0.050 M $$\ce{HCl}$$?
5. Which of the following is usually referred to as strong acid in water solution?

$$\ce{HF}$$,  $$\ce{HNO2}$$,  $$\ce{H2CO3}$$,  $$\ce{H2S}$$,  $$\ce{HSO4-}$$,  $$\ce{Cl-}$$,  $$\ce{HNO3}$$,  $$\ce{HCN}$$

### Solutions

Hint...
You do not need a calculator to evaluate -log (0.01) = 2

Hint...
$$\ce{Ba(OH)2 \rightarrow Ba^2+ + 2 OH-}$$

3. Answer 5.0\times 10^{-13

Hint...
$$\mathrm{[OH^-] = \dfrac{0.80}{40} = 0.02\: M}$$; $$[H^+] = \dfrac{1\times 10^{-14}}{0.02} = 5\times 10^{-13 M}$$. The pH is -12.3.

This solution contains 1.83 g of $$\ce{HCl}$$ per liter. $$\mathrm{[H^+] = 0.050}$$.
5. Answer $$\ce{HNO3}$$