Overview
pH is a logarithmic scale that measures the acidity or basicity of a solution. It's based on the concentration of hydrogen ions (H⁺ or H₃O⁺) in solution.
Definitions
pH
pH = − log [ H + ] = − log [ H 3 O + ] \text{pH} = -\log[\text{H}^+] = -\log[\text{H}_3\text{O}^+] pH = − log [ H + ] = − log [ H 3 O + ] pOH
pOH = − log [ OH − ] \text{pOH} = -\log[\text{OH}^-] pOH = − log [ OH − ] Relationship at 25°C
pH + pOH = 14 \text{pH} + \text{pOH} = 14 pH + pOH = 14 [ H + ] [ OH − ] = K w = 1.0 × 10 − 14 [\text{H}^+][\text{OH}^-] = K_w = 1.0 \times 10^{-14} [ H + ] [ OH − ] = K w = 1.0 × 1 0 − 14 The pH Scale
pH Range Classification 0-6.9 Acidic 7.0 Neutral 7.1-14 Basic (Alkaline)
Common pH Values
Substance pH Stomach acid 1-2 Lemon juice 2.4 Vinegar 2.9 Orange juice 3.5 Tomatoes 4.5 Black coffee 5.0 Milk 6.5 Pure water 7.0 Blood 7.4 Seawater 8.0 Baking soda 8.3 Ammonia solution 11.5 Bleach 12.5 Drain cleaner 14
Calculating pH
From [H⁺]
pH = − log [ H + ] \text{pH} = -\log[\text{H}^+] pH = − log [ H + ] Example: [ H + ] = 3.5 × 10 − 4 [\text{H}^+] = 3.5 \times 10^{-4} [ H + ] = 3.5 × 1 0 − 4
pH = − log ( 3.5 × 10 − 4 ) = 3.46 \text{pH} = -\log(3.5 \times 10^{-4}) = 3.46 pH = − log ( 3.5 × 1 0 − 4 ) = 3.46 Finding [H⁺] from pH
[ H + ] = 10 − pH [\text{H}^+] = 10^{-\text{pH}} [ H + ] = 1 0 − pH Example: pH = 5.20
[ H + ] = 10 − 5.20 = 6.3 × 10 − 6 M [\text{H}^+] = 10^{-5.20} = 6.3 \times 10^{-6} \text{ M} [ H + ] = 1 0 − 5.20 = 6.3 × 1 0 − 6 M Calculating pOH
From [OH⁻]
pOH = − log [ OH − ] \text{pOH} = -\log[\text{OH}^-] pOH = − log [ OH − ] From pH
pOH = 14 − pH \text{pOH} = 14 - \text{pH} pOH = 14 − pH Example: pH = 4.5
pOH = 14 − 4.5 = 9.5 \text{pOH} = 14 - 4.5 = 9.5 pOH = 14 − 4.5 = 9.5 [ OH − ] = 10 − 9.5 = 3.2 × 10 − 10 M [\text{OH}^-] = 10^{-9.5} = 3.2 \times 10^{-10} \text{ M} [ OH − ] = 1 0 − 9.5 = 3.2 × 1 0 − 10 M Converting Between [H⁺] and [OH⁻]
[ H + ] [ OH − ] = 1.0 × 10 − 14 [\text{H}^+][\text{OH}^-] = 1.0 \times 10^{-14} [ H + ] [ OH − ] = 1.0 × 1 0 − 14 Example: [ H + ] = 2.0 × 10 − 3 [\text{H}^+] = 2.0 \times 10^{-3} [ H + ] = 2.0 × 1 0 − 3
[ OH − ] = 1.0 × 10 − 14 2.0 × 10 − 3 = 5.0 × 10 − 12 M [\text{OH}^-] = \frac{1.0 \times 10^{-14}}{2.0 \times 10^{-3}} = 5.0 \times 10^{-12} \text{ M} [ OH − ] = 2.0 × 1 0 − 3 1.0 × 1 0 − 14 = 5.0 × 1 0 − 12 M pKa and pKb
pKa (Acid Strength)
p K a = − log K a \text{p}K_a = -\log K_a p K a = − log K a K a = 10 − p K a K_a = 10^{-\text{p}K_a} K a = 1 0 − p K a Lower pKa = Stronger acid
pKb (Base Strength)
p K b = − log K b \text{p}K_b = -\log K_b p K b = − log K b K b = 10 − p K b K_b = 10^{-\text{p}K_b} K b = 1 0 − p K b Lower pKb = Stronger base
Relationship
p K a + p K b = 14 (for conjugate acid-base pairs) \text{p}K_a + \text{p}K_b = 14 \quad \text{(for conjugate acid-base pairs)} p K a + p K b = 14 (for conjugate acid-base pairs) K a × K b = K w K_a \times K_b = K_w K a × K b = K w Significant Figures in pH
The number of decimal places in pH equals the number of significant figures in [H⁺].
[H⁺] Sig Figs pH 1.0 × 10 − 3 1.0 \times 10^{-3} 1.0 × 1 0 − 3 2 3.00 4.5 × 10 − 6 4.5 \times 10^{-6} 4.5 × 1 0 − 6 2 5.35 2.75 × 10 − 8 2.75 \times 10^{-8} 2.75 × 1 0 − 8 3 7.561
Summary of Key Equations
Find Formula pH − log [ H + ] -\log[\text{H}^+] − log [ H + ] [H⁺] 10 − pH 10^{-\text{pH}} 1 0 − pH pOH − log [ OH − ] -\log[\text{OH}^-] − log [ OH − ] 14 − pH 14 - \text{pH} 14 − pH [OH⁻] 10 − pOH 10^{-\text{pOH}} 1 0 − pOH K w / [ H + ] K_w/[\text{H}^+] K w / [ H + ] K w K_w K w [ H + ] [ OH − ] = 10 − 14 [\text{H}^+][\text{OH}^-] = 10^{-14} [ H + ] [ OH − ] = 1 0 − 14 pH + pOH 14
pH of Strong Acids and Bases
Strong Acids (complete dissociation)
[ H + ] = concentration of acid [\text{H}^+] = \text{concentration of acid} [ H + ] = concentration of acid Example: 0.01 M HCl
[ H + ] = 0.01 M [\text{H}^+] = 0.01 \text{ M} [ H + ] = 0.01 M pH = − log ( 0.01 ) = 2.00 \text{pH} = -\log(0.01) = 2.00 pH = − log ( 0.01 ) = 2.00 Strong Bases (complete dissociation)
[ OH − ] = concentration × number of OH − [\text{OH}^-] = \text{concentration} \times \text{number of OH}^- [ OH − ] = concentration × number of OH − Example: 0.01 M Ba(OH)₂
[ OH − ] = 0.01 × 2 = 0.02 M [\text{OH}^-] = 0.01 \times 2 = 0.02 \text{ M} [ OH − ] = 0.01 × 2 = 0.02 M pOH = − log ( 0.02 ) = 1.70 \text{pOH} = -\log(0.02) = 1.70 pOH = − log ( 0.02 ) = 1.70 pH = 14 − 1.70 = 12.30 \text{pH} = 14 - 1.70 = 12.30 pH = 14 − 1.70 = 12.30 pH of Weak Acids and Bases
Weak Acid
K a = [ H + ] [ A − ] [ HA ] K_a = \frac{[\text{H}^+][\text{A}^-]}{[\text{HA}]} K a = [ HA ] [ H + ] [ A − ] For weak acid HA with concentration C:
[ H + ] = K a × C (if C / K a > 100 ) [\text{H}^+] = \sqrt{K_a \times C} \quad \text{(if } C/K_a > 100\text{)} [ H + ] = K a × C (if C / K a > 100 ) Weak Base
K b = [ BH + ] [ OH − ] [ B ] K_b = \frac{[\text{BH}^+][\text{OH}^-]}{[\text{B}]} K b = [ B ] [ BH + ] [ OH − ] For weak base B with concentration C:
[ OH − ] = K b × C (if C / K b > 100 ) [\text{OH}^-] = \sqrt{K_b \times C} \quad \text{(if } C/K_b > 100\text{)} [ OH − ] = K b × C (if C / K b > 100 ) Temperature Dependence
K w K_w K w
Temperature K w K_w K w pH of neutral water 0°C 1.14 × 10 − 15 1.14 \times 10^{-15} 1.14 × 1 0 − 15 7.47 25°C 1.00 × 10 − 14 1.00 \times 10^{-14} 1.00 × 1 0 − 14 7.00 50°C 5.47 × 10 − 14 5.47 \times 10^{-14} 5.47 × 1 0 − 14 6.63 100°C 5.13 × 10 − 13 5.13 \times 10^{-13} 5.13 × 1 0 − 13 6.14
Note: Neutral always means [ H + ] = [ OH − ] [\text{H}^+] = [\text{OH}^-] [ H + ] = [ OH − ]