5.6 pH or pOH of Acid/Base Aqueous Solutions
The pH of an acid or base solution can be determined by using the Ka, Kb, pKa, or pKb of the acid or base. Solve for the hydronium or hydroxide ion concentration via a typical equilibrium problem (and ICE table) and convert to pH or pOH.
5.6.1 Strong Acid
Example
What is the pH and pOH of a 0.1 M HCl aqueous solution (at 25 °C)?
Ka(HCl) = 2 × 106
Long Version
Strong acids (essentially) dissociate completely to give
\[\mathrm{HA}(aq) \longrightarrow \mathrm{H^+}(aq) + \mathrm{A^-}(aq)\] or with water explicitly in the chemical equation
\[\mathrm{HA}(aq) + \mathrm{H_2O}(l) \longrightarrow \mathrm{H_3O^+}(aq) + \mathrm{A^-}(aq)\]
Therefore, the concentration of the strong acid, [HA], will equal the product concentrations at equilibrium.
\[[\mathrm{HA}]_{\mathrm{i}} = [\mathrm{H^+}]_{\mathrm{eq}}/[\mathrm{H_3O^+}]_{\mathrm{eq}} = [\mathrm{A^-}]_{\mathrm{eq}}\]
To demonstrate, I present the work for this problem below. It is a typical equilibrium problem. Note that water already contains some small amount of H3O+ and OH–. At 25 °C, these concentrations are equal to 1 × 10–7 M.
| HCl(aq) | + | H2O(l) | ⇌ | H3O+(aq) | + | Cl–(aq) | |
|---|---|---|---|---|---|---|---|
I |
0.1 |
1.00E-07 |
0 | ||||
C |
-x |
+x |
+x | ||||
E |
0.1-x |
1.00E-07 + x |
x |
Solve for x in the equilibrium expression.
\[\begin{align*} \dfrac{[\mathrm{H_3O^+}][\mathrm{Cl^-}]}{[\mathrm{HCl}]} &= K \\[1.5ex] \dfrac{(1.00\times 10^{-7} + x)(x)}{0.1-x} &= 2\times 10^{6} \\[1.5ex] 1.00\times 10^{-7}x + x^2 &= 2\times 10^{6} (0.1-x) \\[1.5ex] x^2 + 1.00\times 10^{-7}x &= 2\times 10^{5} - 2\times 10^{6}x\\[1.5ex] x^2 + (2\times 10^{6}x + 1\times 10^{-7}x) - 2\times 10^{5} &= 0 \\[1.5ex] x^2 + 2\times 10^{6}x - 2\times 10^{5} &= 0 \\[1.5ex] x &= 0.1 \end{align*}\]
See quadratic solution at WolframAlpha.
\[\begin{align*} [\mathrm{HCl}]_{\mathrm{eq}} &= 0~M \\ [\mathrm{H_3O^+}]_{\mathrm{eq}} &= 0.1~M\\ [\mathrm{Cl^-}]_{\mathrm{eq}} &= 0.1~M \end{align*}\]
Notice that all the strong acid converted into products.
Solve for the pH and pOH. Note that pKw = 14 at 25 °C.
\[\begin{align*} \mathrm{pH} &= -\log [\mathrm{H_3O^+}] \\[1.5ex] &= -\log (0.1~M) \\[1.5ex] &= 1 \\[1.5ex] \end{align*}\]
\[\begin{align*} \mathrm{pH} + \mathrm{pOH} &= \mathrm{p}K_{\mathrm{w}} \\[1.5ex] \mathrm{pOH} &= \mathrm{p}K_{\mathrm{w}} - \mathrm{pH} \\[1.5ex] &= 14 -1 \\[1.5ex] &= 13 \end{align*}\]
Recognizing that a strong acid completely converts into hydronium
\[[\mathrm{HA}]_{\mathrm{i}} = [\mathrm{H_3O^+}]_{\mathrm{eq}}\] allows one to skip all the work and simply solve for the pH and pOH.
Short Version
Given that the initial H3O+ concentration is very small compared to to the starting acid, we could approximate this concentration to be zero. This cleans up the math/work nicely. I will solve using this approximation to demonstrate.
| HCl(aq) | + | H2O(l) | ⇌ | H3O+(aq) | + | Cl–(aq) | |
|---|---|---|---|---|---|---|---|
I |
0.1 |
~0 |
0 | ||||
C |
-x |
+x |
+x | ||||
E |
0.1-x |
x |
x |
Solve for x in the equilibrium expression.
\[\begin{align*} \dfrac{[\mathrm{H_3O^+}][\mathrm{Cl^-}]}{[\mathrm{HCl}]} &= K_{\mathrm{a}} \\[1.5ex] \dfrac{x^2}{0.1-x} &= 2\times 10^{6} \\[1.5ex] x^2 &= 2\times 10^{6} (0.1-x) \\[1.5ex] &= 2\times 10^{5} - 2\times 10^{6}x\\[1.5ex] x^2 + 2\times 10^{6}x - 2\times 10^{5} &= 0 \\[1.5ex] x &= 0.1 \end{align*}\]
See quadratic solution at WolframAlpha.
\[\begin{align*} [\mathrm{HCl}]_{\mathrm{eq}} &= 0~M \\ [\mathrm{H_3O^+}]_{\mathrm{eq}} &= 0.1~M\\ [\mathrm{Cl^-}]_{\mathrm{eq}} &= 0.1~M \end{align*}\]
Notice that all the strong acid converted into products just as we solved before. This is a fantastic approximation to make.
Solve for the pH and pOH. Note that pKw = 14 at 25 °C.
\[\begin{align*} \mathrm{pH} &= -\log [\mathrm{H_3O^+}] \\[1.5ex] &= -\log (0.1~M) \\[1.5ex] &= 1 \\[1.5ex] \end{align*}\]
\[\begin{align*} \mathrm{pH} + \mathrm{pOH} &= \mathrm{p}K_{\mathrm{w}} \\[1.5ex] \mathrm{pOH} &= \mathrm{p}K_{\mathrm{w}} - \mathrm{pH} \\[1.5ex] &= 14 -1 \\[1.5ex] &= 13 \end{align*}\]
Recognizing that a strong acid completely converts into hydronium
\[[\mathrm{HA}]_{\mathrm{i}} = [\mathrm{H_3O^+}]_{\mathrm{eq}}\] allows one to skip all the work and simply solve for the pH and pOH.
Short Short Version
Recognize that all strong acid converts to hydronium and conjugate base:
0.1 M HCl → 0 M HCl, 0.1 M H3O+, and 0.1 M Cl–
\[\begin{align*} [\mathrm{HCl}]_{\mathrm{eq}} &= 0~M \\ [\mathrm{H_3O^+}]_{\mathrm{eq}} &= 0.1~M\\ [\mathrm{Cl^-}]_{\mathrm{eq}} &= 0.1~M \end{align*}\]
Solve for the pH and pOH. Note that pKw = 14 at 25 °C.
\[\begin{align*} \mathrm{pH} &= -\log [\mathrm{H_3O^+}] \\[1.5ex] &= -\log (0.1~M) \\[1.5ex] &= 1 \\[1.5ex] \end{align*}\]
\[\begin{align*} \mathrm{pH} + \mathrm{pOH} &= \mathrm{p}K_{\mathrm{w}} \\[1.5ex] \mathrm{pOH} &= \mathrm{p}K_{\mathrm{w}} - \mathrm{pH} \\[1.5ex] &= 14 - 1 \\[1.5ex] &= 13 \end{align*}\]
5.6.2 Weak Acid
Example
What is the pH and pOH of a 0.1 M CH3COOH aqueous solution (at 25 °C)?
Ka(CH3COOH) = 1.75 × 10–5
Long Version
Weak acids do not dissociate completely. Determine the hydronium ion concentration (via ICE table in an equilibrium problem) and find the pH and pOH.
| CH3COOH(aq) | + | H2O(l) | ⇌ | H3O+(aq) | + | CH3COO–(aq) | |
|---|---|---|---|---|---|---|---|
I |
0.1 |
1.00E-07 |
0 | ||||
C |
-x |
+x |
+x | ||||
E |
0.1-x |
1.00E-07+x |
x |
\[\begin{align*} \dfrac{[\mathrm{H_3O^+}][\mathrm{CH_3COO^-}]}{[\mathrm{CH_3COOH}]} &= K_{\mathrm{a}} \\[1.5ex] \dfrac{(1\times 10^{-7} + x)(x)}{0.1-x} &= 1.75\times 10^{-5} \\[1.5ex] x^2 + 1\times 10^{-7}x &= 1.75\times 10^{-5}(0.1-x) \\[1.5ex] x^2 + 1\times 10^{-7}x &= 1.75\times 10^{-6} - 1.75\times 10^{-5}x \\[1.5ex] x^2 + (1\times 10^{-7}x + 1.75\times 10^{-5}x) - 1.75\times 10^{-6} &= 0 \\[1.5ex] x^2 + 1.76\times 10^{-5}x - 1.75\times 10^{-6} &= 0 \\[1.5ex] x &= 1.314\times 10^{-3} \end{align*}\]
See quadratic solution at WolframAlpha.
\[\begin{align*} [\mathrm{CH_3COOH}]_{\mathrm{eq}} &= 9.87\times 10^{-2}~M \\ [\mathrm{H_3O^+}]_{\mathrm{eq}} &= 1.31\times 10^{-3}~M\\ [\mathrm{CH_3COO^-}]_{\mathrm{eq}} &= 1.31\times 10^{-3}~M \end{align*}\]
Solve for the pH and pOH. Note that pKw = 14 at 25 °C.
\[\begin{align*} \mathrm{pH} &= -\log [\mathrm{H_3O^+}] \\[1.5ex] &= -\log (1.31\times 10^{-3}~M) \\[1.5ex] &= 2.88 \\[1.5ex] \end{align*}\]
\[\begin{align*} \mathrm{pH} + \mathrm{pOH} &= \mathrm{p}K_{\mathrm{w}} \\[1.5ex] \mathrm{pOH} &= \mathrm{p}K_{\mathrm{w}} - \mathrm{pH} \\[1.5ex] &= 14 - 2.88 \\[1.5ex] &= 11.12 \end{align*}\]
Short Version
Weak acids do not dissociate completely. Determine the hydronium ion concentration (via ICE table in an equilibrium problem) and find the pH and pOH.
| CH3COOH(aq) | + | H2O(l) | ⇌ | H3O+(aq) | + | CH3COO–(aq) | |
|---|---|---|---|---|---|---|---|
I |
0.1 |
~0 |
0 | ||||
C |
-x |
+x |
+x | ||||
E |
0.1-x |
x |
x |
\[\begin{align*} \dfrac{[\mathrm{H_3O^+}][\mathrm{CH_3COO^-}]}{[\mathrm{CH_3COOH}]} &= K_{\mathrm{a}} \\[1.5ex] \dfrac{x^2}{0.1-x} &= 1.75\times 10^{-5} \\[1.5ex] \dfrac{x^2}{0.1} &= 1.75\times 10^{-5} \\[1.5ex] x^2 &= 1.75\times 10^{-6} \\[1.5ex] x &= 1.32\times 10^{-3} \quad (4.18\%) \end{align*}\]
\[\begin{align*} [\mathrm{CH_3COOH}]_{\mathrm{eq}} &= 9.87\times 10^{-2}~M \\ [\mathrm{H_3O^+}]_{\mathrm{eq}} &= 1.32\times 10^{-3}~M\\ [\mathrm{CH_3COO^-}]_{\mathrm{eq}} &= 1.32\times 10^{-3}~M \end{align*}\]
Solve for the pH and pOH. Note that pKw = 14 at 25 °C.
\[\begin{align*} \mathrm{pH} &= -\log [\mathrm{H_3O^+}] \\[1.5ex] &= -\log (1.32\times 10^{-3}~M) \\[1.5ex] &= 2.88 \\[1.5ex] \end{align*}\]
\[\begin{align*} \mathrm{pH} + \mathrm{pOH} &= \mathrm{p}K_{\mathrm{w}} \\[1.5ex] \mathrm{pOH} &= \mathrm{p}K_{\mathrm{w}} - \mathrm{pH} \\[1.5ex] &= 14 - 2.88 \\[1.5ex] &= 11.12 \end{align*}\]
5.6.3 Strong Base
Example
What is the pH and pOH of a 0.1 M NaOH aqueous solution (at 25 °C)?
Short Version
Note that NaOH is a soluble, basic salt. It should (essentially), dissociate completely in water.
\[\mathrm{NaOH}(s) \overset{\mathrm{H_2O}}\longrightarrow \mathrm{Na^+}(aq) + \mathrm{OH^-}(aq)\]
Therefore, the concentration of NaOH will equal the OH– concentration.
\[[\mathrm{NaOH}]_{\mathrm{i}} = [\mathrm{OH^-}]_{\mathrm{eq}} = 0.1~M\]
Since the problem states that the concentration of NaOH is 0.1 M, that means that the hydroxide ion concentration will be 0.1 M (at equilibrium)! Of course, water contains a small amount of OH– (1 × 10–7 at 25 °C). However, 0.1 + 1 × 10–7 ≈ 0.1. Therefore, we will ignore the small contribution of OH– from water.
Note that our solution now contains H2O(l), H3O+(aq), OH–(aq), and Na+(aq). However, Na+ will not be involved in the acid-base reaction between hydronium and hydroxide, so we will ignore it. The proper reaction is given in the ICE table below.
| H2O(l) | ⇌ | H3O+(aq) | + | OH–(aq) | |
|---|---|---|---|---|---|
I |
~0 |
0.1 | |||
C |
+x |
+x | |||
E |
+x |
0.1+x |
Use Kw for pure water to determine the hydronium and hydroxide concentrations.
\[\begin{align*} [\mathrm{H_3O^+}][\mathrm{OH^-}] &= K_{\mathrm{w}} \\[1.5ex] x (0.1+x) &= 1\times 10^{-14} \\[1.5ex] x (0.1) &= 1\times 10^{-14} \\[1.5ex] x &= 1\times 10^{-14} (0.1) \\[1.5ex] x &= 1\times 10^{-15} \quad (1\times 10^{-12}\%) \end{align*}\]
\[\begin{align*} [\mathrm{H_3O^+}]_{\mathrm{eq}} &= 1\times 10^{-15}~M \\ [\mathrm{OH^-}]_{\mathrm{eq}} &= 0.1 + 1\times 10^{-15}\\ &=\approx 0.1~M \end{align*}\]
Since we know that, at equilibrium, the hydroxide ion concentration is equal to the initial concentration of the strong base, we would skip the ICE table stuff. Therefore, we simply solve for the pOH of the solution and convert this to a pH.
Solve for the pH and pOH. Note that pKw = 14 at 25 °C.
\[\begin{align*} \mathrm{pOH} &= -\log [\mathrm{OH^-}] \\[1.5ex] &= -\log (0.1~M) \\[1.5ex] &= 1 \\[1.5ex] \end{align*}\]
\[\begin{align*} \mathrm{pH} + \mathrm{pOH} &= \mathrm{p}K_{\mathrm{w}} \\[1.5ex] \mathrm{pH} &= \mathrm{p}K_{\mathrm{w}} - \mathrm{pOH} \\[1.5ex] &= 14 - 1 \\[1.5ex] &= 13 \end{align*}\]
5.6.4 Weak Base
Example
What is the pH of a 0.1 M NH3 aqueous solution (at 25 °C)?
Kb(NH3) = 1.77 × 10–5
Long Version
Weak bases do not dissociate completely. Determine the hydroxide ion concentration (via ICE table in an equilibrium problem) and find the pOH. We will account for the small amount of hydronium and hydroxide from water for this example.
| NH3(aq) | + | H2O(l) | ⇌ | OH–(aq) | + | NH4+(aq) | |
|---|---|---|---|---|---|---|---|
I |
0.1 |
1.00E-07 |
0 | ||||
C |
-x |
+x |
+x | ||||
E |
0.1-x |
1.00E-07 + x |
x |
\[\begin{align*} \dfrac{[\mathrm{OH^-}][\mathrm{NH_4^+}]}{[\mathrm{NH_3}]} &= K_{\mathrm{b}} \\[1.5ex] \dfrac{(1\times 10^{-7} + x)(x)}{0.1-x} &= 1.77\times 10^{-5} \\[1.5ex] 1\times 10^{-7}x + x^2 &= 1.77\times 10^{-5}(0.1-x) \\[1.5ex] x^2 + 1\times 10^{-7}x &= 1.77\times 10^{-6} - 1.77\times 10^{-5}x \\[1.5ex] x^2 + (1\times 10^{-7}x + 1.77\times 10^{-5}x) - 1.77\times 10^{-6} &= 0 \\[1.5ex] x^2 + 1.78\times 10^{-5}x - 1.77\times 10^{-6} &= 0 \\[1.5ex] x &= 1.32\times 10^{-3} \end{align*}\]
See quadratic solution at WolframAlpha.
\[\begin{align*} [\mathrm{NH_3}]_{\mathrm{eq}} &= 9.87\times 10^{-2}~M \\ [\mathrm{OH^-}]_{\mathrm{eq}} &= 1.32\times 10^{-3}~M \\ [\mathrm{NH_4^+}]_{\mathrm{eq}} &= 1.32\times 10^{-3}~M \end{align*}\]
Solve for the pH and pOH. Note that pKw = 14 at 25 °C.
\[\begin{align*} \mathrm{pOH} &= -\log [\mathrm{OH^-}] \\[1.5ex] &= -\log (1.32\times 10^{-3}~M) \\[1.5ex] &= 2.88 \\[1.5ex] \end{align*}\]
\[\begin{align*} \mathrm{pH} + \mathrm{pOH} &= \mathrm{p}K_{\mathrm{w}} \\[1.5ex] \mathrm{pH} &= \mathrm{p}K_{\mathrm{w}} - \mathrm{pOH} \\[1.5ex] &= 14 - 2.88 \\[1.5ex] &= 11.12 \end{align*}\]
Short Version
In this version, we will approximate the hydroxide ion concentration to be zero.
| NH3(aq) | + | H2O(l) | ⇌ | OH–(aq) | + | NH4+(aq) | |
|---|---|---|---|---|---|---|---|
I |
0.1 |
~0 |
0 | ||||
C |
-x |
+x |
+x | ||||
E |
0.1-x |
x |
x |
\[\begin{align*} \dfrac{[\mathrm{OH^-}][\mathrm{NH_4^+}]}{[\mathrm{NH_3}]} &= K \\[1.5ex] \dfrac{x^2}{0.1-x} &= 1.77\times 10^{-5} \\[1.5ex] \dfrac{x^2}{0.1} &= 1.77\times 10^{-5} \\[1.5ex] x^2 &= 1.77\times 10^{-6} \\[1.5ex] x &= 1.33\times 10^{-3} \quad (1.33\%) \end{align*}\]
\[\begin{align*} [\mathrm{NH_3}]_{\mathrm{eq}} &= 9.87\times 10^{-2}~M \\ [\mathrm{OH^-}]_{\mathrm{eq}} &= 1.33\times 10^{-3}~M\\ [\mathrm{NH_4^+}]_{\mathrm{eq}} &= 1.33\times 10^{-3}~M \end{align*}\]
Notice that the concentrations here are essentially the same as they were in the Full Version of the solution. The approximation works well.
Solve for the pH and pOH. Note that pKw = 14 at 25 °C.
\[\begin{align*} \mathrm{pOH} &= -\log [\mathrm{OH^-}] \\[1.5ex] &= -\log (1.33\times 10^{-3}~M) \\[1.5ex] &= 2.88 \\[1.5ex] \end{align*}\]
\[\begin{align*} \mathrm{pH} + \mathrm{pOH} &= \mathrm{p}K_{\mathrm{w}} \\[1.5ex] \mathrm{pH} &= \mathrm{p}K_{\mathrm{w}} - \mathrm{pOH} \\[1.5ex] &= 14 - 2.88 \\[1.5ex] &= 11.12 \end{align*}\]
Note how our pH and pOH matches exactly (to two decimal places) with the previous work done under the Full Version.