Arrhenius: An acid produces H + \mathrm{H^+} H + O H − \mathrm{OH^-} O H −
Bronsted-Lowry: An acid is a proton (H + \mathrm{H^+} H +
Conjugate pairs: When an acid donates a proton, the remaining species is its conjugate base.
H A + B ⇌ A − + B H + \mathrm{HA} + \mathrm{B \rightleftharpoons \mathrm{A^-}} + \mathrm{BH^+} HA + B ⇌ A − + B H + H A / A − \mathrm{HA/A^-} HA/ A − B / B H + \mathrm{B/BH^+} B/B H +
Example: Identify the conjugate acid-base pairs in:
N H 3 + H 2 O ⇌ N H 4 + + O H − \mathrm{NH_3} + \mathrm{H_2\mathrm{O \rightleftharpoons \mathrm{NH_4^+}}} + \mathrm{OH^-} N H 3 + H 2 O ⇌ N H 4 + + O H − N H 3 / N H 4 + \mathrm{NH_3/\mathrm{NH_4^+}} N H 3 / N H 4 + H 2 O / O H − \mathrm{H_2\mathrm{O/\mathrm{OH^-}}} H 2 O/ O H −
Worked Example 1: Identify the conjugate acid-base pairs in the reaction of H S O 4 − \mathrm{HSO_4^-} HS O 4 − H 2 O \mathrm{H_2\mathrm{O}} H 2 O
H S O 4 − + H 2 O ⇌ S O 4 2 − + H 3 O + \mathrm{HSO_4^-} + \mathrm{H_2\mathrm{O \rightleftharpoons \mathrm{SO_4^{2-}}}} + \mathrm{H_3\mathrm{O^+}} HS O 4 − + H 2 O ⇌ S O 4 2 − + H 3 O + H S O 4 − / S O 4 2 − \mathrm{HSO_4^-/\mathrm{SO_4^{2-}}} HS O 4 − / S O 4 2 − H 2 O / H 3 O + \mathrm{H_2\mathrm{O/\mathrm{H_3\mathrm{O^+}}}} H 2 O/ H 3 O + H S O 4 − \mathrm{HSO_4^-} HS O 4 − H 2 O \mathrm{H_2\mathrm{O}} H 2 O
Strong acids are completely dissociated in aqueous solution.
H C l → H + + C l − \mathrm{HCl} \to \mathrm{H^+} + \mathrm{Cl^-} HCl → H + + C l − Common strong acids: \mathrm{HCl}$$\mathrm{HNO_3}$$\mathrm{H_2\mathrm{SO_4}} (first dissociation), H C l O 4 \mathrm{HClO_4} HCl O 4
Weak acids are partially dissociated in aqueous solution.
C H 3 C O O H ⇌ C H 3 C O O − + H + \mathrm{CH_3\mathrm{COOH \rightleftharpoons \mathrm{CH_3\mathrm{COO^-}}}} + \mathrm{H^+} C H 3 COOH ⇌ C H 3 CO O − + H + Common weak acids: \mathrm{CH_3\mathrm{COOH}}$$\mathrm{H_2\mathrm{CO_3}}$$\mathrm{HF} H 3 P O 4 \mathrm{H_3\mathrm{PO_4}} H 3 P O 4
Property Strong acid Weak acid Dissociation Complete Partial Equilibrium Not established Dynamic equilibrium pH at 0.1 M 1.0 Approximately 2.9 Conductivity High Lower Reaction rate with Mg Faster Slower (same at same [ H + ] [\mathrm{H^+]} [ H + ]
Key misconception: Concentration and strength are independent. A 0.001 M strong acid and a 0.001 M weak acid have the same concentration but different [ H + ] [\mathrm{H^+]} [ H + ]
p H = − log 10 [ H + ] \mathrm{pH} = -\log_{10}[\mathrm{H^+]} pH = − log 10 [ H + ] Where [ H + ] [\mathrm{H^+]} [ H + ]
At 25 ° C 25°C 25° C p H = 7 \mathrm{pH} = 7 pH = 7 p H < 7 \mathrm{pH < 7} pH < 7 p H > 7 \mathrm{pH > 7} pH > 7
Worked Example 2: Find the pH of 0.05 M 0.05 \mathrm{ M} 0.05 M H N O 3 \mathrm{HNO_3} HN O 3
p H = − log 10 ( 0.05 ) = 1.30 \mathrm{pH} = -\log_{10}(0.05) = 1.30 pH = − log 10 ( 0.05 ) = 1.30 Worked Example 3: Find [ H + ] [\mathrm{H^+]} [ H + ]
[ H + ] = 10 − 3.40 = 3.98 × 10 − 4 m o l / L [\mathrm{H^+]} = 10^{-3.40} = 3.98 \times 10^{-4} \mathrm{ mol/L} [ H + ] = 1 0 − 3.40 = 3.98 × 1 0 − 4 mol/L Worked Example 4: Find the pH of 0.005 M 0.005 \mathrm{ M} 0.005 M H 2 S O 4 \mathrm{H_2\mathrm{SO_4}} H 2 S O 4
[ H + ] = 0.005 M [\mathrm{H^+]} = 0.005 \mathrm{ M} [ H + ] = 0.005 M p H = − log 10 ( 0.005 ) = 2.30 \mathrm{pH} = -\log_{10}(0.005) = 2.30 pH = − log 10 ( 0.005 ) = 2.30 Water undergoes autoionisation:
H 2 O ⇌ H + + O H − \mathrm{H_2\mathrm{O \rightleftharpoons \mathrm{H^+}}} + \mathrm{OH^-} H 2 O ⇌ H + + O H − K w = [ H + ] [ O H − ] = 1.0 × 10 − 14 m o l 2 L − 2 a t 25 ° C K_w = [\mathrm{H^+][\mathrm{OH^-]}} = 1.0 \times 10^{-14} \mathrm{ mol^2\mathrm{L^{-2} \quad \mathrm{at 25°C}}} K w = [ H + ] [ O H − ] = 1.0 × 1 0 − 14 mo l 2 L − 2 at25°C Derivation of K w K_w K w
From the autoionisation equilibrium:
K w = [ H + ] [ O H − ] K_w = [\mathrm{H^+][\mathrm{OH^-]}} K w = [ H + ] [ O H − ] In pure water at 25 ° C 25°C 25° C [ H + ] = [ O H − ] = 10 − 7 M [\mathrm{H^+]} = [\mathrm{OH^-]} = 10^{-7} \mathrm{ M} [ H + ] = [ O H − ] = 1 0 − 7 M K w = 10 − 7 × 10 − 7 = 10 − 14 K_w = 10^{-7} \times 10^{-7} = 10^{-14} K w = 1 0 − 7 × 1 0 − 7 = 1 0 − 14
K w K_w K w K w K_w K w [ H + ] = [ O H − ] [\mathrm{H^+]} = [\mathrm{OH^-]} [ H + ] = [ O H − ]
Worked Example 5: Find the pH of 0.02 M 0.02 \mathrm{ M} 0.02 M N a O H \mathrm{NaOH} NaOH
[ O H − ] = 0.02 M [\mathrm{OH^-]} = 0.02 \mathrm{ M} [ O H − ] = 0.02 M [ H + ] = K w [ O H − ] = 1.0 × 10 − 14 0.02 = 5.0 × 10 − 13 M [\mathrm{H^+]} = \frac{K_w}{[\mathrm{OH^-]}} = \frac{1.0 \times 10^{-14}}{0.02} = 5.0 \times 10^{-13} \mathrm{ M} [ H + ] = [ O H − ] K w = 0.02 1.0 × 1 0 − 14 = 5.0 × 1 0 − 13 M p H = − log 10 ( 5.0 × 10 − 13 ) = 12.30 \mathrm{pH} = -\log_{10}(5.0 \times 10^{-13}) = 12.30 pH = − log 10 ( 5.0 × 1 0 − 13 ) = 12.30 For a weak acid H A ⇌ H + + A − \mathrm{HA \rightleftharpoons \mathrm{H^+}} + \mathrm{A^-} HA ⇌ H + + A −
K_a = \frac{[\mathrm{H^+][\mathrm{A^-]}{[\mathrm{HA]} p K a = − log 10 K a pK_a = -\log_{10} K_a p K a = − log 10 K a
The lower the p K a pK_a p K a
Worked Example 6: Ethanoic acid has K a = 1.74 × 10 − 5 m o l / L K_a = 1.74 \times 10^{-5} \mathrm{ mol/L} K a = 1.74 × 1 0 − 5 mol/L 0.10 M 0.10 \mathrm{ M} 0.10 M
K_a = \frac{[\mathrm{H^+][\mathrm{A^-]}{[\mathrm{HA]} = \frac{[\mathrm{H^+]^2}{0.10 - [\mathrm{H^+]} \approx \frac{[\mathrm{H^+]^2}{0.10} [ H + ] = 1.74 × 10 − 5 × 0.10 = 1.74 × 10 − 6 = 1.32 × 10 − 3 M [\mathrm{H^+]} = \sqrt{1.74 \times 10^{-5} \times 0.10} = \sqrt{1.74 \times 10^{-6}} = 1.32 \times 10^{-3} \mathrm{ M} [ H + ] = 1.74 × 1 0 − 5 × 0.10 = 1.74 × 1 0 − 6 = 1.32 × 1 0 − 3 M p H = − log 10 ( 1.32 × 10 − 3 ) = 2.88 \mathrm{pH} = -\log_{10}(1.32 \times 10^{-3}) = 2.88 pH = − log 10 ( 1.32 × 1 0 − 3 ) = 2.88 Worked Example 7: A weak acid H X \mathrm{HX} HX K a = 4.2 × 10 − 4 K_a = 4.2 \times 10^{-4} K a = 4.2 × 1 0 − 4 0.25 M 0.25 \mathrm{ M} 0.25 M
[ H + ] = 4.2 × 10 − 4 × 0.25 = 1.05 × 10 − 4 = 1.025 × 10 − 2 M [\mathrm{H^+]} = \sqrt{4.2 \times 10^{-4} \times 0.25} = \sqrt{1.05 \times 10^{-4}} = 1.025 \times 10^{-2} \mathrm{ M} [ H + ] = 4.2 × 1 0 − 4 × 0.25 = 1.05 × 1 0 − 4 = 1.025 × 1 0 − 2 M p H = − log 10 ( 1.025 × 10 − 2 ) = 1.99 \mathrm{pH} = -\log_{10}(1.025 \times 10^{-2}) = 1.99 pH = − log 10 ( 1.025 × 1 0 − 2 ) = 1.99 % d i s s o c i a t i o n = 1.025 × 10 − 2 0.25 × 100 = 4.1 % \%\mathrm{ dissociation} = \frac{1.025 \times 10^{-2}}{0.25} \times 100 = 4.1\% % dissociation = 0.25 1.025 × 1 0 − 2 × 100 = 4.1% For a weak base B + H 2 O ⇌ B H + + O H − \mathrm{B} + \mathrm{H_2\mathrm{O \rightleftharpoons \mathrm{BH^+}}} + \mathrm{OH^-} B + H 2 O ⇌ B H + + O H −
K_b = \frac{[\mathrm{BH^+][\mathrm{OH^-]}{[\mathrm{B]} Relationship:
K a × K b = K w K_a \times K_b = K_w K a × K b = K w Proof: For a conjugate pair H A / A − \mathrm{HA/A^-} HA/ A −
K_a = \frac{[\mathrm{H^+][\mathrm{A^-]}{[\mathrm{HA]} \quad \mathrm{and \quad K_b = \frac{[\mathrm{HA][\mathrm{OH^-]}{[\mathrm{A^-]} K_a \times K_b = \frac{[\mathrm{H^+][\mathrm{A^-]}{[\mathrm{HA]} \times \frac{[\mathrm{HA][\mathrm{OH^-]}{[\mathrm{A^-]} = [\mathrm{H^+][\mathrm{OH^-] = K_w Worked Example 8: Ammonia has K b = 1.78 × 10 − 5 m o l / L K_b = 1.78 \times 10^{-5} \mathrm{ mol/L} K b = 1.78 × 1 0 − 5 mol/L 0.15 M 0.15 \mathrm{ M} 0.15 M
[ O H − ] = K b × [ B ] = 1.78 × 10 − 5 × 0.15 = 2.67 × 10 − 6 = 1.63 × 10 − 3 M [\mathrm{OH^-]} = \sqrt{K_b \times [\mathrm{B]}} = \sqrt{1.78 \times 10^{-5} \times 0.15} = \sqrt{2.67 \times 10^{-6}} = 1.63 \times 10^{-3} \mathrm{ M} [ O H − ] = K b × [ B ] = 1.78 × 1 0 − 5 × 0.15 = 2.67 × 1 0 − 6 = 1.63 × 1 0 − 3 M p O H = − log 10 ( 1.63 × 10 − 3 ) = 2.79 \mathrm{pOH} = -\log_{10}(1.63 \times 10^{-3}) = 2.79 pOH = − log 10 ( 1.63 × 1 0 − 3 ) = 2.79 p H = 14 − 2.79 = 11.21 \mathrm{pH} = 14 - 2.79 = 11.21 pH = 14 − 2.79 = 11.21 A buffer solution resists changes in pH when small amounts of acid or base are added. It consists of a weak acid and its conjugate base (or a weak base and its conjugate acid).
Acidic buffer: Weak acid (H A \mathrm{HA} HA A − \mathrm{A^-} A −
Example: Ethanoic acid + sodium ethanoate.
Basic buffer: Weak base (B \mathrm{B} B B H + \mathrm{BH^+} B H +
Example: Ammonia + ammonium chloride.
\mathrm{pH = pK_a + \log_{10}\left(\frac{[\mathrm{A^-]}{[\mathrm{HA]}\right) Derivation:
Starting from the acid dissociation expression:
K_a = \frac{[\mathrm{H^+][\mathrm{A^-]}{[\mathrm{HA]} Rearranging: [\mathrm{H^+] = K_a \times \frac{[\mathrm{HA]}{[\mathrm{A^-]}
Taking − log 10 -\log_{10} − log 10
-\log[\mathrm{H^+] = -\log K_a - \log\frac{[\mathrm{HA]}{[\mathrm{A^-]} \mathrm{pH = pK_a + \log\frac{[\mathrm{A^-]}{[\mathrm{HA]} Worked Example 9: Calculate the pH of a buffer containing 0.20 M 0.20 \mathrm{ M} 0.20 M p K a = 4.76 pK_a = 4.76 p K a = 4.76 0.15 M 0.15 \mathrm{ M} 0.15 M
p H = 4.76 + log 10 ( 0.15 0.20 ) = 4.76 + log 10 ( 0.75 ) = 4.76 − 0.125 = 4.64 \mathrm{pH} = 4.76 + \log_{10}\left(\frac{0.15}{0.20}\right) = 4.76 + \log_{10}(0.75) = 4.76 - 0.125 = 4.64 pH = 4.76 + log 10 ( 0.20 0.15 ) = 4.76 + log 10 ( 0.75 ) = 4.76 − 0.125 = 4.64 Worked Example 10: Prepare a buffer at pH 5.00 using ethanoic acid (p K a = 4.76 pK_a = 4.76 p K a = 4.76 0.30 M 0.30 \mathrm{ M} 0.30 M
5.00 = 4.76 + \log\frac{[\mathrm{A^-]}{[\mathrm{HA]} log [ A − ] [ H A ] = 0.24 \log\frac{[\mathrm{A^-]}{[\mathrm{HA]}}} = 0.24 log = [ A − ] [ HA ] 0.24 [ A − ] [ H A ] = 10 0.24 = 1.74 \frac{[\mathrm{A^-]}{[\mathrm{HA]}}} = 10^{0.24} = 1.74 = [ A − ] [ HA ] 1 0 0.24 = 1.74 Let [ H A ] = x [\mathrm{HA]} = x [ HA ] = x [ A − ] = 1.74 x [\mathrm{A^-]} = 1.74x [ A − ] = 1.74 x
x + 1.74 x = 0.30 ⟹ 2.74 x = 0.30 ⟹ x = 0.109 M x + 1.74x = 0.30 \implies 2.74x = 0.30 \implies x = 0.109 \mathrm{ M} x + 1.74 x = 0.30 ⟹ 2.74 x = 0.30 ⟹ x = 0.109 M
[ H A ] = 0.109 M [\mathrm{HA]} = 0.109 \mathrm{ M} [ HA ] = 0.109 M [ A − ] = 0.191 M [\mathrm{A^-]} = 0.191 \mathrm{ M} [ A − ] = 0.191 M
The buffer capacity depends on:
The absolute concentrations of the weak acid and conjugate base (higher concentrations = greater capacity) The ratio [ A − ] / [ H A ] [\mathrm{A^-]/[\mathrm{HA]}} [ A − ] / [ HA ] p K a pK_a p K a Worked Example 11: A buffer contains 0.10 M 0.10 \mathrm{ M} 0.10 M N H 3 \mathrm{NH_3} N H 3 0.15 M 0.15 \mathrm{ M} 0.15 M N H 4 C l \mathrm{NH_4\mathrm{Cl}} N H 4 Cl 0.01 m o l 0.01 \mathrm{ mol} 0.01 mol H C l \mathrm{HCl} HCl 1 L 1 \mathrm{ L} 1 L
First, find K a K_a K a N H 4 + \mathrm{NH_4^+} N H 4 +
K a = K w K b = 1.0 × 10 − 14 1.78 × 10 − 5 = 5.62 × 10 − 10 K_a = \frac{K_w}{K_b} = \frac{1.0 \times 10^{-14}}{1.78 \times 10^{-5}} = 5.62 \times 10^{-10} K a = K b K w = 1.78 × 1 0 − 5 1.0 × 1 0 − 14 = 5.62 × 1 0 − 10 p K a = 9.25 pK_a = 9.25 p K a = 9.25 p H = 9.25 + log 0.10 0.15 = 9.25 − 0.176 = 9.07 \mathrm{pH} = 9.25 + \log\frac{0.10}{0.15} = 9.25 - 0.176 = 9.07 pH = 9.25 + log 0.15 0.10 = 9.25 − 0.176 = 9.07 After adding 0.01 m o l 0.01 \mathrm{ mol} 0.01 mol H C l \mathrm{HCl} HCl
N H 3 \mathrm{NH_3} N H 3 H + \mathrm{H^+} H + [ N H 3 ] [\mathrm{NH_3]} [ N H 3 ] 0.01 0.01 0.01 [ N H 4 + ] [\mathrm{NH_4^+]} [ N H 4 + ] 0.01 0.01 0.01
[ N H 3 ] = 0.10 − 0.01 = 0.09 M [\mathrm{NH_3]} = 0.10 - 0.01 = 0.09 \mathrm{ M} [ N H 3 ] = 0.10 − 0.01 = 0.09 M [ N H 4 + ] = 0.15 + 0.01 = 0.16 M [\mathrm{NH_4^+]} = 0.15 + 0.01 = 0.16 \mathrm{ M} [ N H 4 + ] = 0.15 + 0.01 = 0.16 M p H = 9.25 + log 0.09 0.16 = 9.25 + log ( 0.5625 ) = 9.25 − 0.250 = 9.00 \mathrm{pH} = 9.25 + \log\frac{0.09}{0.16} = 9.25 + \log(0.5625) = 9.25 - 0.250 = 9.00 pH = 9.25 + log 0.16 0.09 = 9.25 + log ( 0.5625 ) = 9.25 − 0.250 = 9.00 The pH changes by only 0.07 units, demonstrating the buffer”s effectiveness.
Equivalence point at pH 7.
Worked Example 12: 25.0 c m 3 25.0 \mathrm{ cm^3} 25.0 c m 3 0.10 M 0.10 \mathrm{ M} 0.10 M H C l \mathrm{HCl} HCl 0.10 M 0.10 \mathrm{ M} 0.10 M N a O H \mathrm{NaOH} NaOH
At the equivalence point: moles of acid = moles of base.
n = 0.10 × 0.0250 = 0.00250 m o l n = 0.10 \times 0.0250 = 0.00250 \mathrm{ mol} n = 0.10 × 0.0250 = 0.00250 mol Total volume = 50.0 c m 3 50.0 \mathrm{ cm^3} 50.0 c m 3
[ N a C l ] = 0.00250 / 0.0500 = 0.0500 M [\mathrm{NaCl]} = 0.00250/0.0500 = 0.0500 \mathrm{ M} [ NaCl ] = 0.00250/0.0500 = 0.0500 M
p H = 7 \mathrm{pH} = 7 pH = 7 Equivalence point at pH < 7 (acidic).
Equivalence point at pH > 7 (alkaline).
Worked Example 13: 25.0 c m 3 25.0 \mathrm{ cm^3} 25.0 c m 3 0.10 M 0.10 \mathrm{ M} 0.10 M C H 3 C O O H \mathrm{CH_3\mathrm{COOH}} C H 3 COOH K a = 1.74 × 10 − 5 K_a = 1.74 \times 10^{-5} K a = 1.74 × 1 0 − 5 0.10 M 0.10 \mathrm{ M} 0.10 M N a O H \mathrm{NaOH} NaOH
Moles of C H 3 C O O − \mathrm{CH_3\mathrm{COO^-}} C H 3 CO O − 0.00250 m o l 0.00250 \mathrm{ mol} 0.00250 mol
Total volume = 50.0 c m 3 50.0 \mathrm{ cm^3} 50.0 c m 3
[ C H 3 C O O − ] = 0.0500 M [\mathrm{CH_3\mathrm{COO^-]}} = 0.0500 \mathrm{ M} [ C H 3 CO O − ] = 0.0500 M
The ethanoate ion hydrolyses:
C H 3 C O O − + H 2 O ⇌ C H 3 C O O H + O H − \mathrm{CH_3\mathrm{COO^-}} + \mathrm{H_2\mathrm{O \rightleftharpoons \mathrm{CH_3\mathrm{COOH}}}} + \mathrm{OH^-} C H 3 CO O − + H 2 O ⇌ C H 3 COOH + O H − K b = K w K a = 1.0 × 10 − 14 1.74 × 10 − 5 = 5.75 × 10 − 10 K_b = \frac{K_w}{K_a} = \frac{1.0 \times 10^{-14}}{1.74 \times 10^{-5}} = 5.75 \times 10^{-10} K b = K a K w = 1.74 × 1 0 − 5 1.0 × 1 0 − 14 = 5.75 × 1 0 − 10 [ O H − ] = K b × [ C H 3 C O O − ] = 5.75 × 10 − 10 × 0.0500 = 2.875 × 10 − 11 = 5.36 × 10 − 6 M [\mathrm{OH^-]} = \sqrt{K_b \times [\mathrm{CH_3\mathrm{COO^-]}}} = \sqrt{5.75 \times 10^{-10} \times 0.0500} = \sqrt{2.875 \times 10^{-11}} = 5.36 \times 10^{-6} \mathrm{ M} [ O H − ] = K b × [ C H 3 CO O − ] = 5.75 × 1 0 − 10 × 0.0500 = 2.875 × 1 0 − 11 = 5.36 × 1 0 − 6 M p O H = − log 10 ( 5.36 × 10 − 6 ) = 5.27 \mathrm{pOH} = -\log_{10}(5.36 \times 10^{-6}) = 5.27 pOH = − log 10 ( 5.36 × 1 0 − 6 ) = 5.27 p H = 14 − 5.27 = 8.73 \mathrm{pH} = 14 - 5.27 = 8.73 pH = 14 − 5.27 = 8.73 Worked Example 14: 20.0 c m 3 20.0 \mathrm{ cm^3} 20.0 c m 3 0.15 M 0.15 \mathrm{ M} 0.15 M N H 3 \mathrm{NH_3} N H 3 K b = 1.78 × 10 − 5 K_b = 1.78 \times 10^{-5} K b = 1.78 × 1 0 − 5 0.10 M 0.10 \mathrm{ M} 0.10 M H C l \mathrm{HCl} HCl 15.0 c m 3 15.0 \mathrm{ cm^3} 15.0 c m 3 H C l \mathrm{HCl} HCl
Moles of N H 3 = 0.15 × 0.0200 = 0.00300 m o l \mathrm{NH_3} = 0.15 \times 0.0200 = 0.00300 \mathrm{ mol} N H 3 = 0.15 × 0.0200 = 0.00300 mol H C l \mathrm{HCl} HCl = 0.10 × 0.0150 = 0.00150 m o l = 0.10 \times 0.0150 = 0.00150 \mathrm{ mol} = 0.10 × 0.0150 = 0.00150 mol
After reaction: [ N H 3 ] [\mathrm{NH_3]} [ N H 3 ] = 0.00300 − 0.00150 = 0.00150 m o l = 0.00300 - 0.00150 = 0.00150 \mathrm{ mol} = 0.00300 − 0.00150 = 0.00150 mol [ N H 4 + ] [\mathrm{NH_4^+]} [ N H 4 + ] = 0.00150 m o l = 0.00150 \mathrm{ mol} = 0.00150 mol
Total volume = 35.0 c m 3 = 0.0350 L = 35.0 \mathrm{ cm^3} = 0.0350 \mathrm{ L} = 35.0 c m 3 = 0.0350 L
[ N H 3 ] = 0.00150 / 0.0350 = 0.0429 M [\mathrm{NH_3]} = 0.00150/0.0350 = 0.0429 \mathrm{ M} [ N H 3 ] = 0.00150/0.0350 = 0.0429 M [ N H 4 + ] = 0.00150 / 0.0350 = 0.0429 M [\mathrm{NH_4^+]} = 0.00150/0.0350 = 0.0429 \mathrm{ M} [ N H 4 + ] = 0.00150/0.0350 = 0.0429 M
This is a buffer solution with equal concentrations, so:
p H = p K a + log [ N H 3 ] [ N H 4 + ] = 9.25 + log ( 1 ) = 9.25 \mathrm{pH} = pK_a + \log\frac{[\mathrm{NH_3]}{[\mathrm{NH_4^+]}}} = 9.25 + \log(1) = 9.25 pH = p K a + log = [ N H 3 ] [ N H 4 + ] 9.25 + log ( 1 ) = 9.25 An indicator is a weak acid where H I n \mathrm{HIn} HIn I n − \mathrm{In^-} I n −
H I n ⇌ H + + I n − \mathrm{HIn \rightleftharpoons \mathrm{H^+}} + \mathrm{In^-} HIn ⇌ H + + I n − The indicator changes colour over approximately p K I n ± 1 \mathrm{pK_{\mathrm{In} \pm 1}} p K In ± 1
Indicator pH Range Colour Change Methyl orange 3.1 - 4.4 Red to yellow Bromothymol blue 6.0 - 7.6 Yellow to blue Phenolphthalein 8.3 - 10.0 Colourless to pink
The indicator range must overlap with the steep part of the titration curve at the equivalence Point.
Titration type Equivalence pH Suitable indicator Strong acid + strong base pH = 7 Bromothymol blue Strong acid + weak base pH < 7 Methyl orange Weak acid + strong base pH > 7 Phenolphthalein
For a strong acid-weak base titration” — this is incorrect. Phenolphthalein (pH 8.3-10.0) is suitable for weak acid-strong base titrations where the equivalence pH is above 7. For strong Acid-weak base titrations (equivalence pH below 7), methyl orange is the correct choice.
Acids that can donate two protons, e.g., H 2 S O 4 \mathrm{H_2\mathrm{SO_4}} H 2 S O 4 H 2 C O 3 \mathrm{H_2\mathrm{CO_3}} H 2 C O 3 H 3 P O 4 \mathrm{H_3\mathrm{PO_4}} H 3 P O 4
For carbonic acid:
H 2 C O 3 ⇌ H + + H C O 3 − K a 1 = 4.3 × 10 − 7 \mathrm{H_2\mathrm{CO_3 \rightleftharpoons \mathrm{H^+}}} + \mathrm{HCO_3^- \quad K_{a1}} = 4.3 \times 10^{-7} H 2 C O 3 ⇌ H + + HC O 3 − K a1 = 4.3 × 1 0 − 7 H C O 3 − ⇌ H + + C O 3 2 − K a 2 = 4.8 × 10 − 11 \mathrm{HCO_3^- \rightleftharpoons \mathrm{H^+}} + \mathrm{CO_3^{2-} \quad K_{a2}} = 4.8 \times 10^{-11} HC O 3 − ⇌ H + + C O 3 2 − K a2 = 4.8 × 1 0 − 11 Note that K a 1 ≫ K a 2 K_{a1} \gg K_{a2} K a 1 ≫ K a 2
Concept Formula/Definition Key Point pH p H = − log [ H + ] \mathrm{pH} = -\log[\mathrm{H^+]} pH = − log [ H + ] Lower pH = more acidic K w K_w K w [ H + ] [ O H − ] = 10 − 14 [\mathrm{H^+][\mathrm{OH^-]}} = 10^{-14} [ H + ] [ O H − ] = 1 0 − 14 At 25°C K a K_a K a \frac{[\mathrm{H^+][\mathrm{A^-]}{[\mathrm{HA]} Higher K a K_a K a K b K_b K b \frac{[\mathrm{BH^+][\mathrm{OH^-]}{[\mathrm{B]} Higher K b K_b K b Henderson-Hasselbalch \mathrm{pH = pK_a + \log\frac{[\mathrm{A^-]}{[\mathrm{HA]} For buffer solutions K a × K b K_a \times K_b K a × K b = K w = K_w = K w For conjugate pairs
Strong vs. Weak acids: A strong acid is completely dissociated; a weak acid is partially dissociated. Concentration and strength are independent.
[ H + ] [\mathrm{H^+]} [ H + ] [ H + ] = K a × c [\mathrm{H^+]} = \sqrt{K_a \times c} [ H + ] = K a × c [ H + ] [\mathrm{H^+]} [ H + ] c c c c / K a > 100 c/K_a > 100 c / K a > 100
pH of water: Pure water has pH 7 at 25 ° C 25°C 25° C K w K_w K w
Buffer calculations: The Henderson-Hasselbalch equation uses concentrations, not moles (unless the acid and base are in the same volume).
Choosing an indicator: The indicator range must include the pH at the equivalence point.
Conjugate base strength: The weaker the acid, the stronger its conjugate base. C l − \mathrm{Cl^-} C l − H C l \mathrm{HCl} HCl
Diprotic acids: The second dissociation is always weaker than the first. For H 2 S O 4 \mathrm{H_2\mathrm{SO_4}} H 2 S O 4 K a 2 = 1.2 × 10 − 2 K_{a2} = 1.2 \times 10^{-2} K a 2 = 1.2 × 1 0 − 2
Calculate the pH of 0.005 M 0.005 \mathrm{ M} 0.005 M H 2 S O 4 \mathrm{H_2\mathrm{SO_4}} H 2 S O 4
A weak acid H X \mathrm{HX} HX K a = 4.2 × 10 − 4 K_a = 4.2 \times 10^{-4} K a = 4.2 × 1 0 − 4 0.25 M 0.25 \mathrm{ M} 0.25 M
Prepare a buffer at pH 5.00 using ethanoic acid (p K a = 4.76 pK_a = 4.76 p K a = 4.76 0.30 M 0.30 \mathrm{ M} 0.30 M
20.0 c m 3 20.0 \mathrm{ cm^3} 20.0 c m 3 0.15 M 0.15 \mathrm{ M} 0.15 M N H 3 \mathrm{NH_3} N H 3 K b = 1.78 × 10 − 5 K_b = 1.78 \times 10^{-5} K b = 1.78 × 1 0 − 5 0.10 M 0.10 \mathrm{ M} 0.10 M H C l \mathrm{HCl} HCl 15.0 c m 3 15.0 \mathrm{ cm^3} 15.0 c m 3 H C l \mathrm{HCl} HCl
Explain why the pH of a 0.01 M 0.01 \mathrm{ M} 0.01 M H C l \mathrm{HCl} HCl 0.01 M 0.01 \mathrm{ M} 0.01 M C H 3 C O O H \mathrm{CH_3\mathrm{COOH}} C H 3 COOH
A buffer contains 0.10 M 0.10 \mathrm{ M} 0.10 M N H 3 \mathrm{NH_3} N H 3 0.15 M 0.15 \mathrm{ M} 0.15 M N H 4 C l \mathrm{NH_4\mathrm{Cl}} N H 4 Cl 0.01 m o l 0.01 \mathrm{ mol} 0.01 mol H C l \mathrm{HCl} HCl 1 L 1 \mathrm{ L} 1 L
Calculate K a K_a K a 0.050 M 0.050 \mathrm{ M} 0.050 M
Explain why phenolphthalein is a suitable indicator for a weak acid-strong base titration but methyl orange is suitable for a strong acid-weak base titration.
Calculate the pH of a 0.15 M 0.15 \mathrm{ M} 0.15 M H F \mathrm{HF} HF K a = 6.8 × 10 − 4 K_a = 6.8 \times 10^{-4} K a = 6.8 × 1 0 − 4 [ H + ] ≈ K a × c [\mathrm{H^+] \approx \sqrt{K_a \times c}} [ H + ] ≈ K a × c
50 mL of 0.10 M 0.10 \mathrm{ M} 0.10 M N a O H \mathrm{NaOH} NaOH 0.10 M 0.10 \mathrm{ M} 0.10 M C H 3 C O O H \mathrm{CH_3\mathrm{COOH}} C H 3 COOH
A student prepares a buffer by mixing 100 m L 100 \mathrm{ mL} 100 mL 0.20 M 0.20 \mathrm{ M} 0.20 M 50 m L 50 \mathrm{ mL} 50 mL 0.20 M 0.20 \mathrm{ M} 0.20 M N a O H \mathrm{NaOH} NaOH
Explain why the pH of a solution of N a C l \mathrm{NaCl} NaCl N H 4 C l \mathrm{NH_4\mathrm{Cl}} N H 4 Cl
When a salt dissolves in water, its ions may react with water (hydrolyse), affecting the pH of the solution.
Example: N a C l \mathrm{NaCl} NaCl K N O 3 \mathrm{KNO_3} KN O 3
Neither ion reacts with water. The solution is neutral (pH = 7).
Example: N H 4 C l \mathrm{NH_4\mathrm{Cl}} N H 4 Cl
The cation (N H 4 + \mathrm{NH_4^+} N H 4 +
N H 4 + + H 2 O ⇌ N H 3 + H 3 O + \mathrm{NH_4^+} + \mathrm{H_2\mathrm{O \rightleftharpoons \mathrm{NH_3}}} + \mathrm{H_3\mathrm{O^+}} N H 4 + + H 2 O ⇌ N H 3 + H 3 O + This produces H + \mathrm{H^+} H +
Worked Example 15: Calculate the pH of a 0.10 M 0.10 \mathrm{ M} 0.10 M N H 4 C l \mathrm{NH_4\mathrm{Cl}} N H 4 Cl K b ( N H 3 ) = 1.78 × 10 − 5 K_b(\mathrm{NH_3)} = 1.78 \times 10^{-5} K b ( N H 3 ) = 1.78 × 1 0 − 5
K a ( N H 4 + ) = K w K b = 1.0 × 10 − 14 1.78 × 10 − 5 = 5.62 × 10 − 10 K_a(\mathrm{NH_4^+)} = \frac{K_w}{K_b} = \frac{1.0 \times 10^{-14}}{1.78 \times 10^{-5}} = 5.62 \times 10^{-10} K a ( N H 4 + ) = K b K w = 1.78 × 1 0 − 5 1.0 × 1 0 − 14 = 5.62 × 1 0 − 10 [ H + ] = K a × c = 5.62 × 10 − 10 × 0.10 = 5.62 × 10 − 11 = 7.50 × 10 − 6 M [\mathrm{H^+]} = \sqrt{K_a \times c} = \sqrt{5.62 \times 10^{-10} \times 0.10} = \sqrt{5.62 \times 10^{-11}} = 7.50 \times 10^{-6} \mathrm{ M} [ H + ] = K a × c = 5.62 × 1 0 − 10 × 0.10 = 5.62 × 1 0 − 11 = 7.50 × 1 0 − 6 M p H = − log ( 7.50 × 10 − 6 ) = 5.12 \mathrm{pH} = -\log(7.50 \times 10^{-6}) = 5.12 pH = − log ( 7.50 × 1 0 − 6 ) = 5.12 Example: C H 3 C O O N a \mathrm{CH_3\mathrm{COONa}} C H 3 COONa
The anion (C H 3 C O O − \mathrm{CH_3\mathrm{COO^-}} C H 3 CO O −
C H 3 C O O − + H 2 O ⇌ C H 3 C O O H + O H − \mathrm{CH_3\mathrm{COO^-}} + \mathrm{H_2\mathrm{O \rightleftharpoons \mathrm{CH_3\mathrm{COOH}}}} + \mathrm{OH^-} C H 3 CO O − + H 2 O ⇌ C H 3 COOH + O H − This produces O H − \mathrm{OH^-} O H −
Example: N H 4 C H 3 C O O \mathrm{NH_4\mathrm{CH_3\mathrm{COO}}} N H 4 C H 3 COO
Both ions hydrolyse. The pH depends on the relative strengths of the acid and base:
If K a > K b K_a > K_b K a > K b If K a < K b K_a < K_b K a < K b If K a ≈ K b K_a \approx K_b K a ≈ K b Salt type Hydrolysis pH Example Strong acid + strong base Neither ion hydrolyses 7 N a C l \mathrm{NaCl} NaCl Strong acid + weak base Cation hydrolyses < 7 N H 4 C l \mathrm{NH_4\mathrm{Cl}} N H 4 Cl Weak acid + strong base Anion hydrolyses > 7 C H 3 C O O N a \mathrm{CH_3\mathrm{COONa}} C H 3 COONa Weak acid + weak base Both hydrolyse Depends N H 4 C H 3 C O O \mathrm{NH_4\mathrm{CH_3\mathrm{COO}}} N H 4 C H 3 COO
[ H + ] t o t a l = n 1 + n 2 V t o t a l [\mathrm{H^+]_{\mathrm{total}}} = \frac{n_1 + n_2}{V_{\mathrm{total}}} [ H + ] total = V total n 1 + n 2 Determine which is in excess, then calculate [ H + ] [\mathrm{H^+]} [ H + ] [ O H − ] [\mathrm{OH^-]} [ O H − ]
Worked Example 16: 50 m L 50 \mathrm{ mL} 50 mL 0.10 M 0.10 \mathrm{ M} 0.10 M N a O H \mathrm{NaOH} NaOH 50 m L 50 \mathrm{ mL} 50 mL 0.10 M 0.10 \mathrm{ M} 0.10 M C H 3 C O O H \mathrm{CH_3\mathrm{COOH}} C H 3 COOH
Moles of N a O H = 0.10 × 0.050 = 0.0050 m o l \mathrm{NaOH} = 0.10 \times 0.050 = 0.0050 \mathrm{ mol} NaOH = 0.10 × 0.050 = 0.0050 mol C H 3 C O O H = 0.10 × 0.050 = 0.0050 m o l \mathrm{CH_3\mathrm{COOH}} = 0.10 \times 0.050 = 0.0050 \mathrm{ mol} C H 3 COOH = 0.10 × 0.050 = 0.0050 mol
This is the equivalence point: all C H 3 C O O H \mathrm{CH_3\mathrm{COOH}} C H 3 COOH C H 3 C O O − \mathrm{CH_3\mathrm{COO^-}} C H 3 CO O −
[ C H 3 C O O − ] = 0.0050 / 0.100 = 0.050 M [\mathrm{CH_3\mathrm{COO^-]}} = 0.0050/0.100 = 0.050 \mathrm{ M} [ C H 3 CO O − ] = 0.0050/0.100 = 0.050 M
K b = K w K a = 1.0 × 10 − 14 1.74 × 10 − 5 = 5.75 × 10 − 10 K_b = \frac{K_w}{K_a} = \frac{1.0 \times 10^{-14}}{1.74 \times 10^{-5}} = 5.75 \times 10^{-10} K b = K a K w = 1.74 × 1 0 − 5 1.0 × 1 0 − 14 = 5.75 × 1 0 − 10 [ O H − ] = 5.75 × 10 − 10 × 0.050 = 5.36 × 10 − 6 M [\mathrm{OH^-]} = \sqrt{5.75 \times 10^{-10} \times 0.050} = 5.36 \times 10^{-6} \mathrm{ M} [ O H − ] = 5.75 × 1 0 − 10 × 0.050 = 5.36 × 1 0 − 6 M p O H = 5.27 , p H = 8.73 \mathrm{pOH} = 5.27, \quad \mathrm{pH} = 8.73 pOH = 5.27 , pH = 8.73 If both acids are present, the strong acid dominates the pH because it contributes far more H + \mathrm{H^+} H +
Strong acid-strong base: The pH changes slowly at first, then rapidly near the equivalence point (almost vertical), then slowly again. The equivalence point is at pH 7.
Weak acid-strong base: The initial pH is higher than for a strong acid at the same Concentration. There is a buffering region in the middle of the curve (where the weak acid and its conjugate base coexist). The equivalence point is at pH > 7.
Strong acid-weak base: The equivalence point is at pH < 7.
Worked Example 17: Sketch and describe the titration curve for 25 m L 25 \mathrm{ mL} 25 mL 0.10 M 0.10 \mathrm{ M} 0.10 M C H 3 C O O H \mathrm{CH_3\mathrm{COOH}} C H 3 COOH 0.10 M 0.10 \mathrm{ M} 0.10 M N a O H \mathrm{NaOH} NaOH
Key features:
Initial pH: ~2.87 (weak acid)Buffer region: From ~0 to ~20 mL added. The pH changes slowly because the solution acts as a buffer (C H 3 C O O H / C H 3 C O O − \mathrm{CH_3\mathrm{COOH/\mathrm{CH_3\mathrm{COO^-}}}} C H 3 COOH/ C H 3 CO O − Half-equivalence point: At 12.5 m L 12.5 \mathrm{ mL} 12.5 mL p H = p K a = 4.76 \mathrm{pH} = pK_a = 4.76 pH = p K a = 4.76 Equivalence point: At 25 m L 25 \mathrm{ mL} 25 mL Beyond equivalence: pH approaches that of the excess N a O H \mathrm{NaOH} NaOH The solubility of some salts is affected by pH.
C a C O 3 \mathrm{CaCO_3} CaC O 3 C O 3 2 − \mathrm{CO_3^{2-}} C O 3 2 − H + \mathrm{H^+} H +
C O 3 2 − + H + → H C O 3 − \mathrm{CO_3^{2-}} + \mathrm{H^+} \to \mathrm{HCO_3^-} C O 3 2 − + H + → HC O 3 − H C O 3 − + H + → H 2 C O 3 → C O 2 + H 2 O \mathrm{HCO_3^-} + \mathrm{H^+} \to \mathrm{H_2\mathrm{CO_3}} \to \mathrm{CO_2} + \mathrm{H_2\mathrm{O}} HC O 3 − + H + → H 2 C O 3 → C O 2 + H 2 O This removes C O 3 2 − \mathrm{CO_3^{2-}} C O 3 2 −
M g ( O H ) 2 \mathrm{Mg(OH)_2} Mg ( OH ) 2 O H − \mathrm{OH^-} O H − H + \mathrm{H^+} H +
M g ( O H ) 2 ( s ) + 2 H + → M g 2 + + 2 H 2 O \mathrm{Mg(OH)_2\mathrm{(s)}} + 2\mathrm{H^+} \to \mathrm{Mg^{2+}} + 2\mathrm{H_2\mathrm{O}} Mg ( OH ) 2 ( s ) + 2 H + → M g 2 + + 2 H 2 O M g ( O H ) 2 \mathrm{Mg(OH)_2} Mg ( OH ) 2 O H − \mathrm{OH^-} O H −
Calculate the pH of a 0.050 M 0.050 \mathrm{ M} 0.050 M A l C l 3 \mathrm{AlCl_3} AlC l 3 K a ( A l 3 + ) = 1.0 × 10 − 5 K_a(\mathrm{Al^{3+})} = 1.0 \times 10^{-5} K a ( A l 3 + ) = 1.0 × 1 0 − 5 A l 3 + \mathrm{Al^{3+}} A l 3 +
A buffer is prepared by mixing 0.20 m o l 0.20 \mathrm{ mol} 0.20 mol C H 3 C O O H \mathrm{CH_3\mathrm{COOH}} C H 3 COOH 0.10 m o l 0.10 \mathrm{ mol} 0.10 mol N a O H \mathrm{NaOH} NaOH 1.0 L 1.0 \mathrm{ L} 1.0 L
Explain qualitatively how the shape of the pH titration curve for a weak acid-strong base titration differs from that of a strong acid-strong base titration. Include a discussion of the buffer region and the half-equivalence point.
Calculate the pH at each of the following points in the titration of 25.0 m L 25.0 \mathrm{ mL} 25.0 mL 0.10 M 0.10 \mathrm{ M} 0.10 M H C l \mathrm{HCl} HCl 0.10 M 0.10 \mathrm{ M} 0.10 M N a O H \mathrm{NaOH} NaOH
Explain why N a 2 C O 3 \mathrm{Na_2\mathrm{CO_3}} N a 2 C O 3
A student is asked to prepare a buffer at pH 7.40 (physiological pH) using H 2 P O 4 − \mathrm{H_2\mathrm{PO_4^-}} H 2 P O 4 − H P O 4 2 − \mathrm{HPO_4^{2-}} HP O 4 2 − p K a 2 = 7.21 pK_{a2} = 7.21 p K a 2 = 7.21 [ H P O 4 2 − ] / [ H 2 P O 4 − ] [\mathrm{HPO_4^{2-}]/[\mathrm{H_2\mathrm{PO_4^-]}}} [ HP O 4 2 − ] / [ H 2 P O 4 − ]
When the concentration of a strong acid is very low (comparable to 10 − 7 10^{-7} 1 0 − 7 H + \mathrm{H^+} H +
Worked Example 19: Find the pH of 1.0 × 10 − 8 1.0 \times 10^{-8} 1.0 × 1 0 − 8
If we use p H = − log ( 10 − 8 ) = 8 \mathrm{pH} = -\log(10^{-8}) = 8 pH = − log ( 1 0 − 8 ) = 8
[ H + ] t o t a l = [ H + ] H C l + [ H + ] w a t e r = 1.0 × 10 − 8 + [ O H − ] [\mathrm{H^+]_{\mathrm{total}}} = [\mathrm{H^+]_{\mathrm{HCl}}} + [\mathrm{H^+]_{\mathrm{water}}} = 1.0 \times 10^{-8} + [\mathrm{OH^-]} [ H + ] total = [ H + ] HCl + [ H + ] water = 1.0 × 1 0 − 8 + [ O H − ] From K w = [ H + ] [ O H − ] K_w = [\mathrm{H^+][\mathrm{OH^-]}} K w = [ H + ] [ O H − ]
[ H + ] × ( [ H + ] − 1.0 × 10 − 8 ) = 1.0 × 10 − 14 [\mathrm{H^+] \times ([\mathrm{H^+]}} - 1.0 \times 10^{-8}) = 1.0 \times 10^{-14} [ H + ] × ([ H + ] − 1.0 × 1 0 − 8 ) = 1.0 × 1 0 − 14 [ H + ] 2 − 1.0 × 10 − 8 [ H + ] − 1.0 × 10 − 14 = 0 [\mathrm{H^+]^2} - 1.0 \times 10^{-8}[\mathrm{H^+]} - 1.0 \times 10^{-14} = 0 [ H + ] 2 − 1.0 × 1 0 − 8 [ H + ] − 1.0 × 1 0 − 14 = 0 Using the quadratic formula:
[ H + ] = 1.0 × 10 − 8 + ( 1.0 × 10 − 8 ) 2 + 4 × 10 − 14 2 [\mathrm{H^+]} = \frac{1.0 \times 10^{-8} + \sqrt{(1.0 \times 10^{-8})^2 + 4 \times 10^{-14}}}{2} [ H + ] = 2 1.0 × 1 0 − 8 + ( 1.0 × 1 0 − 8 ) 2 + 4 × 1 0 − 14 = 1.0 × 10 − 8 + 10 − 16 + 4 × 10 − 14 2 = 1.0 × 10 − 8 + 4.01 × 10 − 14 2 = \frac{1.0 \times 10^{-8} + \sqrt{10^{-16} + 4 \times 10^{-14}}}{2} = \frac{1.0 \times 10^{-8} + \sqrt{4.01 \times 10^{-14}}}{2} = 2 1.0 × 1 0 − 8 + 1 0 − 16 + 4 × 1 0 − 14 = 2 1.0 × 1 0 − 8 + 4.01 × 1 0 − 14 = 1.0 × 10 − 8 + 2.0025 × 10 − 7 2 = 2.1025 × 10 − 7 2 = 1.051 × 10 − 7 M = \frac{1.0 \times 10^{-8} + 2.0025 \times 10^{-7}}{2} = \frac{2.1025 \times 10^{-7}}{2} = 1.051 \times 10^{-7} \mathrm{ M} = 2 1.0 × 1 0 − 8 + 2.0025 × 1 0 − 7 = 2 2.1025 × 1 0 − 7 = 1.051 × 1 0 − 7 M p H = − log ( 1.051 × 10 − 7 ) = 6.98 \mathrm{pH} = -\log(1.051 \times 10^{-7}) = 6.98 pH = − log ( 1.051 × 1 0 − 7 ) = 6.98 This makes sense: the pH is very slightly below 7, consistent with a very dilute acid.
For H 3 P O 4 \mathrm{H_3\mathrm{PO_4}} H 3 P O 4
K a 1 = 7.5 × 10 − 3 , K a 2 = 6.2 × 10 − 8 , K a 3 = 4.8 × 10 − 13 K_{a1} = 7.5 \times 10^{-3}, \quad K_{a2} = 6.2 \times 10^{-8}, \quad K_{a3} = 4.8 \times 10^{-13} K a 1 = 7.5 × 1 0 − 3 , K a 2 = 6.2 × 1 0 − 8 , K a 3 = 4.8 × 1 0 − 13 Since K a 1 ≫ K a 2 ≫ K a 3 K_{a1} \gg K_{a2} \gg K_{a3} K a 1 ≫ K a 2 ≫ K a 3 0.10 M 0.10 \mathrm{ M} 0.10 M K a 1 K_{a1} K a 1
An indicator is a weak acid H I n \mathrm{HIn} HIn
H I n ⇌ H + + I n − \mathrm{HIn \rightleftharpoons \mathrm{H^+}} + \mathrm{In^-} HIn ⇌ H + + I n − K_{\mathrm{In} = \frac{[\mathrm{H^+][\mathrm{In^-]}{[\mathrm{HIn]} The human eye sees the acid colour when [ H I n ] / [ I n − ] > 10 [\mathrm{HIn]/[\mathrm{In^-] \gt 10}} [ HIn ] / [ I n − ] > 10 [ I n − ] / [ H I n ] > 10 [\mathrm{In^-]/[\mathrm{HIn] \gt 10}} [ I n − ] / [ HIn ] > 10 p K I n ± 1 \mathrm{pK_{\mathrm{In} \pm 1}} p K In ± 1
Example 1: pH calculation
Calculate the pH of a 0.050 mol dm − 3 0.050\,\text{mol\,dm}^{-3} 0.050 mol dm − 3
Solution:
HCl is a strong acid, so [ H + ] = 0.050 mol dm − 3 [\text{H}^+] = 0.050\,\text{mol\,dm}^{-3} [ H + ] = 0.050 mol dm − 3
pH = − log 10 [ H + ] = − log 10 ( 0.050 ) = 1.30 \text{pH} = -\log_{10}[\text{H}^+] = -\log_{10}(0.050) = 1.30 pH = − log 10 [ H + ] = − log 10 ( 0.050 ) = 1.30 Chemistry explains how atoms combine to form the substances that make up everything around you. Bonding is about how electrons are shared or transferred between atoms, determining properties like melting point and conductivity. Chemical reactions are rearrangements of atoms, where old bonds break and new ones form. The mole concept bridges the atomic world and the laboratory, letting you predict exactly how much product a reaction will produce. Understanding these principles lets you predict the behavior of matter before you even enter the lab.
